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Section 5.11 Using Function Graphs to Determine Solution Sets to Inequalities

When solving inequalities involving functions that are presented in graphical form, we follow a two-step process. We first use the function to identify the points on the curve whose \(-y\)-coordinates satisfy the property implied by the inequality statement. We then identify the \(x\)-coordinates of those points, which collectively make up the solution set to the inequality.

In FigureĀ 5.11.1, I've indicated all of the points on the function named \(g\) that have \(y\)-coordinates greater than or equal to \(1\text{.}\) I've also marked off the portion of the \(x\)-axis over which these points lie. Since the \(y\)-coordinates of the points are the values of \(g(x)\text{,}\) we can infer from this that the solution set to the inequality \(g(x) \geq 1\) is \([-4.5,3]\text{.}\)

The graph of a function named \(g\text{.}\) The function resembles an upside down vee. The function resembles an upside down vee. There is an arrow at the end of the left side of the vee and a solid dot t the end of the right half of the vee which terminates at the point \((5,-1)\text{.}\) The vertex of the vee is at the point \((-2,6)\text{.}\) The dashed horizontal line \(y=1\) is also graphed. The left side of the vee intersects the line at the point \((-5,1)\) and the right side of the vee intersects the line at the point \((3,1)\text{.}\) There are solid dots at each of the points of intersection. The portion of the vee that lies between the two points of intersection is highlighted. The interval \([-5,3]\) is highlighted on the \(x\)-axis.
Figure 5.11.1. \(y=g(x)\)

In FigureĀ 5.11.2, I've indicated all of the points on the function named \(g\) that have \(y\)-coordinates less than to \(44\text{.}\) I've also marked off the portion of the \(x\)-axis over which these points lie. Since the \(y\)-coordinates of the points are the values of \(g(x)\text{,}\) we can infer from this that the solution set to the inequality \(g(x) \lt 4\) is \((-\infty,-3) \cup (0,6]\text{.}\)

The graph of a function named \(g\text{.}\) The function resembles an upside down vee. There is an arrow at the end of the left side of the vee and a solid dot t the end of the right half of the vee which terminates at the point \((5,-1)\text{.}\) The vertex of the vee is at the point \((-2,6)\text{.}\) The dashed horizontal line \(y=4\) is also graphed. The left side of the vee intersects the line at the point \((-3,4)\) and the right side of the vee intersects the line at the point \((0,4)\text{.}\) There are open circles at each point of intersection. The portion of the vee that lies to the left of the point \((-3,4)\) is highlighted as is the portion of the vee that lies to the right of the point \((0,4)\text{.}\) The interval \((-\infty,-3) \cup (0,5]\) is highlighted on the \(x\)-axis.
Figure 5.11.2. \(y=g(x)\)

You can use FigureĀ 5.11.3 and FigureĀ 5.11.4 to investigate these idea in an interactive manner.

Figure 5.11.3. The parabola is the function \(f\text{.}\) Drag the button labeled "up/down" - umm... - up and down to change the problem and solution.
Figure 5.11.4. The parabola is the function \(f\text{.}\) Drag the button labeled "up/down" - umm... - up and down to change the problem and solution.

Exercises Exercises

Determine the solution set to each stated inequality.

1.

Determine the solution set to \(f(x) \lt 2\) based upon the function \(f\) shown in FigureĀ 5.11.5. State the solution set using both set-builder notation and interval notation.

The graph of a function named f. The function is rather squiggly. The left most point on the function is a solid dot at the point \((-4,2)\text{.}\) The right most point plotted is an open circle at the point \((5,5)\text{.}\) From left to right, the curve starts at the point \((-4,2)\) and rises to a peak at \((-3,5)\text{.}\) It then falls past the point \((-2,2)\) until it reaches a floor at the point \((-1,-1)\text{.}\) It then rises past the point \((0,2)\) until reaching a second peak at the point \((1,5)\text{.}\) The curve then falls past the point \((2,2)\) to a second floor at the point \((3,-1)\text{.}\) The curve finishes up by rising past the point \((4,1)\) before terminating at the hole plotted at the point \((5,5)\text{.}\)
Figure 5.11.5. \(y=f(x)\)

Solution

The solution set is \(\{x \mid -2 \lt x \lt 0 \text{ or } 2 \lt x \lt 4\}\text{.}\)

The solution set is \((-2,0) \cup (2,4)\text{.}\)

The graph of a function named f. The function is rather squiggly. The left most point on the function is a solid dot at the point \((-4,2)\text{.}\) The right most point plotted is an open circle at the point \((5,5)\text{.}\) From left to right, the curve starts at the point \((-4,2)\) and rises to a peak at \((-3,5)\text{.}\) It then falls past the point \((-2,2)\) until it reaches a floor at the point \((-1,-1)\text{.}\) It then rises past the point \((0,2)\) until reaching a second peak at the point \((1,5)\text{.}\) The curve then falls past the point \((2,2)\) to a second floor at the point \((3,-1)\text{.}\) The curve finishes up by rising past the point \((4,1)\) before terminating at the hole plotted at the point \((5,5)\text{.}\) The dashed horizontal line \(y=2\) is also graphed. The function intersects the line at the points \((-4,4)\text{,}\) \((-2,4)\text{,}\) \((0,4)\text{,}\) \((2,4)\text{,}\) and \((4,4)\text{.}\) The portions of the curve that fall over the intervals \((-2,0)\) and \((2,4)\) are highlighted with open circles at each endpoint - these are the intervals where the function lies below the line. The intervals \((-2,0)\) and \((2,4)\) are highlighted on the \(y\)-axis.
Figure 5.11.6. \(f(x) \lt 2\)
2.

Determine the solution set to \(f(x) \geq 2\) based upon the function \(f\) shown in FigureĀ 5.11.7. State the solution set using both set-builder notation and interval notation.

The graph of a function named f. The function is rather squiggly. The left most point on the function is a solid dot at the point \((-4,2)\text{.}\) The right most point plotted is an open circle at the point \((5,5)\text{.}\) From left to right, the curve starts at the point \((-4,2)\) and rises to a peak at \((-3,5)\text{.}\) It then falls past the point \((-2,2)\) until it reaches a floor at the point \((-1,-1)\text{.}\) It then rises past the point \((0,2)\) until reaching a second peak at the point \((1,5)\text{.}\) The curve then falls past the point \((2,2)\) to a second floor at the point \((3,-1)\text{.}\) The curve finishes up by rising past the point \((4,1)\) before terminating at the hole plotted at the point \((5,5)\text{.}\)
Figure 5.11.7. \(y=f(x)\)

Solution

The solution set is \(\{x \mid -4 \leq x \leq -2 \text{ or } 0 \leq x \leq 2 \text{ or } 4 \leq x \lt 5\}\text{.}\)

The solution set is \([-4,-2] \cup [0,2] \cup [4,5)\text{.}\)

The graph of a function named f. The function is rather squiggly. The left most point on the function is a solid dot at the point \((-4,2)\text{.}\) The right most point plotted is an open circle at the point \((5,5)\text{.}\) From left to right, the curve starts at the point \((-4,2)\) and rises to a peak at \((-3,5)\text{.}\) It then falls past the point \((-2,2)\) until it reaches a floor at the point \((-1,-1)\text{.}\) It then rises past the point \((0,2)\) until reaching a second peak at the point \((1,5)\text{.}\) The curve then falls past the point \((2,2)\) to a second floor at the point \((3,-1)\text{.}\) The curve finishes up by rising past the point \((4,1)\) before terminating at the hole plotted at the point \((5,5)\text{.}\) The dashed horizontal line \(y=2\) is also graphed. The function intersects the line at the points \((-4,4)\text{,}\) \((-2,4)\text{,}\) \((0,4)\text{,}\) \((2,4)\text{,}\) and \((4,4)\text{.}\) The portions of the curve that fall over the intervals \([-4,-2]\text{,}\) \([0,2]\text{,}\) and \([4,5)\) are highlighted with solid dots at each end except for the point \((5,5)\) at which there is an open circle - these are the intervals where the function lies above or on the line. The intervals \([-4,-2]\text{,}\) \([0,2]\text{,}\) and \([4,5)\) are all highlighted on the \(x\)-axis.
Figure 5.11.8. \(f(x) \ge 2\)
3.

Determine the solution set to \(f(x) \lt -1\) based upon the function \(f\) shown in FigureĀ 5.11.9. State the solution set using both set-builder notation and interval notation.

The graph of a function named f. The function is rather squiggly. The left most point on the function is a solid dot at the point \((-4,2)\text{.}\) The right most point plotted is an open circle at the point \((5,5)\text{.}\) From left to right, the curve starts at the point \((-4,2)\) and rises to a peak at \((-3,5)\text{.}\) It then falls past the point \((-2,2)\) until it reaches a floor at the point \((-1,-1)\text{.}\) It then rises past the point \((0,2)\) until reaching a second peak at the point \((1,5)\text{.}\) The curve then falls past the point \((2,2)\) to a second floor at the point \((3,-1)\text{.}\) The curve finishes up by rising past the point \((4,1)\) before terminating at the hole plotted at the point \((5,5)\text{.}\)
Figure 5.11.9. \(y=f(x)\)

Solution

The solution set is \(\{\}\text{.}\)

The solution set is \(\emptyset\text{.}\)

The graph of a function named f. The function is rather squiggly. The left most point on the function is a solid dot at the point \((-4,2)\text{.}\) The right most point plotted is an open circle at the point \((5,5)\text{.}\) From left to right, the curve starts at the point \((-4,2)\) and rises to a peak at \((-3,5)\text{.}\) It then falls past the point \((-2,2)\) until it reaches a floor at the point \((-1,-1)\text{.}\) It then rises past the point \((0,2)\) until reaching a second peak at the point \((1,5)\text{.}\) The curve then falls past the point \((2,2)\) to a second floor at the point \((3,-1)\text{.}\) The curve finishes up by rising past the point \((4,1)\) before terminating at the hole plotted at the point \((5,5)\text{.}\) The dashed horizontal line \(y=-1\) is also drawn. The function intersects the line at the points \((-1,-1)\) and \((3,-1)\) but every other point on the function falls above the line.
Figure 5.11.10. \(f(x) \lt -1\)

4.

Determine the solution set to \(f(x) \leq 5\) based upon the function \(f\) shown in FigureĀ 5.11.11. State the solution set using both set-builder notation and interval notation.

The graph of a function named f. The function is rather squiggly. The left most point on the function is a solid dot at the point \((-4,2)\text{.}\) The right most point plotted is an open circle at the point \((5,5)\text{.}\) From left to right, the curve starts at the point \((-4,2)\) and rises to a peak at \((-3,5)\text{.}\) It then falls past the point \((-2,2)\) until it reaches a floor at the point \((-1,-1)\text{.}\) It then rises past the point \((0,2)\) until reaching a second peak at the point \((1,5)\text{.}\) The curve then falls past the point \((2,2)\) to a second floor at the point \((3,-1)\text{.}\) The curve finishes up by rising past the point \((4,1)\) before terminating at the hole plotted at the point \((5,5)\text{.}\)
Figure 5.11.11. \(y=f(x)\)

Solution

The solution set is \(\{x \mid -4 \leq x \lt 5\}\text{.}\)

The solution set is \((-4,5)\text{.}\)

The graph of a function named f. The function is rather squiggly. The left most point on the function is a solid dot at the point \((-4,2)\text{.}\) The right most point plotted is an open circle at the point \((5,5)\text{.}\) From left to right, the curve starts at the point \((-4,2)\) and rises to a peak at \((-3,5)\text{.}\) It then falls past the point \((-2,2)\) until it reaches a floor at the point \((-1,-1)\text{.}\) It then rises past the point \((0,2)\) until reaching a second peak at the point \((1,5)\text{.}\) The curve then falls past the point \((2,2)\) to a second floor at the point \((3,-1)\text{.}\) The curve finishes up by rising past the point \((4,1)\) before terminating at the hole plotted at the point \((5,5)\text{.}\) The dashed horizontal line \(y=5\) is also drawn. The function intersects the lie at the points \((-3,5)\) and \((1,5)\text{.}\) Every other point on the function lies below the line. The entire function is highlighted and the interval \([-4,5)\) is highlighted on the \(x\)axis.
Figure 5.11.12. \(f(x) \le 5\)
5.

Determine the solution set to \(g(x) \geq -2\) based upon the function \(g\) shown in FigureĀ 5.11.13. State the solution set using both set-builder notation and interval notation.

The graph of a function named g. There are two distinct pieces to the function. The piece on the left resembles a fishhook. It has open holes at either end. The endpoints are \((-6,2)\) and \((-1,3)\text{.}\) The bottom point on the hook is \((4,-6)\text{.}\) The hook also passes through the point \((-2,-2)\text{.}\) The other piece of the function is a half-line. The half-line originates at a solid dot plotted at the point \((-1,6)\text{.}\) The line then decreases with a slope of \(-2\) passes through the points \((0,4)\text{,}\) \((1,2)\text{,}\) \((2,0)\text{,}\) \((3,-2)\text{,}\) \((4,-4)\) and \((5,-6)\text{.}\) There is an arrow at the right end of the half-line.
Figure 5.11.13. \(y=g(x)\)

Solution

The solution set is \(\{-2 \leq x \leq 3\}\text{.}\)

The solution set is \([-2,3]\text{.}\)

The graph of a function named g. There are two distinct pieces to the function. The piece on the left resembles a fishhook. It has open holes at either end. The endpoints are \((-6,2)\) and \((-1,3)\text{.}\) The bottom point on the hook is \((4,-6)\text{.}\) The hook also passes through the point \((-2,-2)\text{.}\) The other piece of the function is a half-line. The half-line originates at a solid dot plotted at the point \((-1,6)\text{.}\) The line then decreases with a slope of \(-2\) passes through the points \((0,4)\text{,}\) \((1,2)\text{,}\) \((2,0)\text{,}\) \((3,-2)\text{,}\) \((4,-4)\) and \((5,-6)\text{.}\) There is an arrow at the right end of the half-line. The dashed horizontal line \(y=-2\) is also graphed. The line intersects the function at the points \((-2,-2)\) and \((3,2)\text{.}\) Solid dots have been placed at each of these points. The part of the function that lies above the line falls between these two points, and that part of the function has been highlighted, The interval \([-2,3]\) has been highlighted on the \(x\)-axis.
Figure 5.11.14. \(g(x) \ge -2\)
6.

Determine the solution set to \(g(x) \leq 3\) based upon the function \(g\) shown in FigureĀ 5.11.15. State the solution set using both set-builder notation and interval notation.

The graph of a function named g. There are two distinct pieces to the function. The piece on the left resembles a fishhook. It has open holes at either end. The endpoints are \((-6,2)\) and \((-1,3)\text{.}\) The bottom point on the hook is \((4,-6)\text{.}\) The hook also passes through the point \((-2,-2)\text{.}\) The other piece of the function is a half-line. The half-line originates at a solid dot plotted at the point \((-1,6)\text{.}\) The line then decreases with a slope of \(-2\) passes through the points \((0,4)\text{,}\) \((1,2)\text{,}\) \((2,0)\text{,}\) \((3,-2)\text{,}\) \((4,-4)\) and \((5,-6)\text{.}\) There is an arrow at the right end of the half-line.
Figure 5.11.15. \(y=g(x)\)

Solution

The solution set is \(\{x \mid -6 \lt x \lt 2 \text{ or } x \geq \frac{1}{2}\}\text{.}\)

The solution set is \((-6,-1) \cup \left[\frac{1}{2},\infty\right)\text{.}\)

The graph of a function named g. There are two distinct pieces to the function. The piece on the left resembles a fishhook. It has open holes at either end. The endpoints are \((-6,2)\) and \((-1,3)\text{.}\) The bottom point on the hook is \((4,-6)\text{.}\) The hook also passes through the point \((-2,-2)\text{.}\) The other piece of the function is a half-line. The half-line originates at a solid dot plotted at the point \((-1,6)\text{.}\) The line then decreases with a slope of \(-2\) passes through the points \((0,4)\text{,}\) \((1,2)\text{,}\) \((2,0)\text{,}\) \((3,-2)\text{,}\) \((4,-4)\) and \((5,-6)\text{.}\) There is an arrow at the right end of the half-line. The dashed horizontal line \(y=3\) is also plotted. The function intersects the line at the point \((0.5,3)\) and a solid dot has been drawn there. The portions of the function that fall over the intervals \((-6,-1)\) and \((0.5,\infty)\) lie below the line and have been highlighted. The interval \((-6,-1) \cup [0.5,\infty)\) has been highlighted on the \(x\)-axis.
Figure 5.11.16. \(g(x) \le 3\)
7.

Determine the solution set to \(g(x) \lt -6\) based upon the function \(g\) shown in FigureĀ 5.11.17. State the solution set using both set-builder notation and interval notation.

The graph of a function named g. There are two distinct pieces to the function. The piece on the left resembles a fishhook. It has open holes at either end. The endpoints are \((-6,2)\) and \((-1,3)\text{.}\) The bottom point on the hook is \((4,-6)\text{.}\) The hook also passes through the point \((-2,-2)\text{.}\) The other piece of the function is a half-line. The half-line originates at a solid dot plotted at the point \((-1,6)\text{.}\) The line then decreases with a slope of \(-2\) passes through the points \((0,4)\text{,}\) \((1,2)\text{,}\) \((2,0)\text{,}\) \((3,-2)\text{,}\) \((4,-4)\) and \((5,-6)\text{.}\) There is an arrow at the right end of the half-line.
Figure 5.11.17. \(y=g(x)\)

Solution

The solution set is \(\{x \mid x \gt 5\}\text{.}\)

The solution set is \((5,\infty)\text{.}\)

The graph of a function named g. There are two distinct pieces to the function. The piece on the left resembles a fishhook. It has open holes at either end. The endpoints are \((-6,2)\) and \((-1,3)\text{.}\) The bottom point on the hook is \((4,-6)\text{.}\) The hook also passes through the point \((-2,-2)\text{.}\) The other piece of the function is a half-line. The half-line originates at a solid dot plotted at the point \((-1,6)\text{.}\) The line then decreases with a slope of \(-2\) passes through the points \((0,4)\text{,}\) \((1,2)\text{,}\) \((2,0)\text{,}\) \((3,-2)\text{,}\) \((4,-4)\) and \((5,-6)\text{.}\) There is an arrow at the right end of the half-line. The dashed horizontal line \(y-6\) is also graphed. The function intersects the line at the points \((-4,-6)\) and \((5,-6)\text{.}\) The only portion of the function that lies below the line is the portion to the right of the point \((5,6)\text{.}\) The interval \((5,\infty)\) is highlighted on the \(x\)-axis.
Figure 5.11.18. \(g(x) \lt -6\)

8.

Determine the solution set to \(g(x) \gt -6\) based upon the function \(g\) shown in FigureĀ 5.11.19. State the solution set using both set-builder notation and interval notation.

The graph of a function named g. There are two distinct pieces to the function. The piece on the left resembles a fishhook. It has open holes at either end. The endpoints are \((-6,2)\) and \((-1,3)\text{.}\) The bottom point on the hook is \((4,-6)\text{.}\) The hook also passes through the point \((-2,-2)\text{.}\) The other piece of the function is a half-line. The half-line originates at a solid dot plotted at the point \((-1,6)\text{.}\) The line then decreases with a slope of \(-2\) passes through the points \((0,4)\text{,}\) \((1,2)\text{,}\) \((2,0)\text{,}\) \((3,-2)\text{,}\) \((4,-4)\) and \((5,-6)\text{.}\) There is an arrow at the right end of the half-line.
Figure 5.11.19. \(y=g(x)\)

Solution

The solution set is \(\{x \mid -6 \lt x \lt -4 \text{ or } -4 \lt x \lt 5\}\text{.}\)

The solution set is \((-6,-4) \cup (-4,5)\text{.}\)

The graph of a function named g. There are two distinct pieces to the function. The piece on the left resembles a fishhook. It has open holes at either end. The endpoints are \((-6,2)\) and \((-1,3)\text{.}\) The bottom point on the hook is \((4,-6)\text{.}\) The hook also passes through the point \((-2,-2)\text{.}\) The other piece of the function is a half-line. The half-line originates at a solid dot plotted at the point \((-1,6)\text{.}\) The line then decreases with a slope of \(-2\) passes through the points \((0,4)\text{,}\) \((1,2)\text{,}\) \((2,0)\text{,}\) \((3,-2)\text{,}\) \((4,-4)\) and \((5,-6)\text{.}\) There is an arrow at the right end of the half-line. The dashed horizontal line \(y-6\) is also graphed. The function intersects the line at the points \((-4,-6)\) and \((5,-6)\text{.}\) The only potion of the function that doesn't lie above the line are those two points and the portion that lies to the right of the point \((5,-6)\text{.}\) The interval \((-6,-4) \cup (-4,5)\) is highlighted on the \(x\)-axis.
Figure 5.11.20. \(g(x) \gt -6\)
9.

Determine the solution set to \(f(x) \geq g(x)\) based upon the functions \(f\) (piecewise-linear) and \(g\) (parabolic) shown in FigureĀ 5.11.21. State the solution set using both set-builder notation and interval notation.

The graph of two functions, one named \(f\) and the other named \(g\text{.}\) The function \(f\) is a downward opening vee with a vertex at the point \((2,5)\text{.}\) The function \(g\) is an upward opening parabola with a vertex at the point \((-3,-4)\text{.}\) The two functions intersect at the points \((-5,-2)\) and \((1,4)\text{.}\)
Figure 5.11.21. \(\highlightr{y=f(x)}\) and \(\highlighty{y=g(x)}\)

Solution

The solution set is \(\{x \mid -5 \leq x \leq 1\}\text{.}\)

The solution set is \([-5,1]\text{.}\)

The graph of two functions, one named \(f\) and the other named \(g\text{.}\) The function \(f\) is a downward opening vee with a vertex at the point \((2,5)\text{.}\) The function \(g\) is an upward opening parabola with a vertex at the point \((-3,-4)\text{.}\) The two functions intersect at the points \((-5,-2)\) and \((1,4)\text{.}\) The function \(f\) lies above the function \(r\) over the interval \([-5,1]\) and this portion of the function \(f\) has been highlighted with solid dots at each end. The interval \([-5,1]\) is highlighted on the \(x\)-axis.
Figure 5.11.22. \(\highlightr{f(x)} \ge \highlighty{g(x)}\)
10.

Determine the solution set to \(g(x) \gt f(x)\) based upon the functions \(f\) (piecewise-linear) and \(g\) (parabolic) shown in FigureĀ 5.11.23. State the solution set using both set-builder notation and interval notation.

The graph of two functions, one named \(f\) and the other named \(g\text{.}\) The function \(f\) is a downward opening vee with a vertex at the point \((2,5)\text{.}\) The function \(g\) is an upward opening parabola with a vertex at the point \((-3,-4)\text{.}\) The two functions intersect at the points \((-5,-2)\) and \((1,4)\text{.}\)
Figure 5.11.23. \(\highlightr{y=f(x)}\) and \(\highlighty{y=g(x)}\)

Solution

The solution set is \(\{x \mid x \lt -5 \text{ or } x \gt 1\}\text{.}\)

The solution set is \((-\infty,-5) \cup (1,\infty)\text{.}\)

The graph of two functions, one named \(f\) and the other named \(g\text{.}\) The function \(f\) is a downward opening vee with a vertex at the point \((2,5)\text{.}\) The function \(g\) is an upward opening parabola with a vertex at the point \((-3,-4)\text{.}\) The two functions intersect at the points \((-5,-2)\) and \((1,4)\text{.}\) The function \(g\) lies above the function \(f\) over the intervals \((-\infty,-5)\) and \((1,\infty)\text{.}\) The portion of \(g\) has been highlighted with open circles at the endpoints. The interval \((-\infty,-5) \cup (1,\infty)\) is highlighted on the \(x\)-axis.
Figure 5.11.24. \(\highlighty{g(x)} \gt \highlightr{f(x)}\)
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