Question 2.5.13: Estimate the solutions of |0.14x²-13.72|>|0.58x|+11...

Precalculus Functions and Graphs [1893897]

Estimate the solutions of

|0.14x²-13.72|>|0.58x|+11

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Equation: |0.14x²-13.72|>|0.58x|+11

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Step 1:

To solve the inequality, we first assign the functions Y1 and Y2 as ABS(0.14x^2-13.72) and ABS(0.58x)+11, respectively. Our goal is to find the values of x for which Y1 is greater than Y2.

Step 2:

To begin, we choose a viewing rectangle of [-30,30] by [0,40] and graph the functions. Since there is symmetry with respect to the y-axis, we only need to find the x-coordinates of the points of intersection for x>0.

Step 3:

Using an intersect feature, we find that the graphs intersect at approximately x=2.80 and x=15.52. Referring to the graph, we can see that the solution to the inequality is the union of three intervals: (-∞,-15.52), (-2.80,2.80), and (15.52,∞).

Step 4:

Therefore, the solution to the inequality ABS(0.14x^2-13.72) > ABS(0.58x)+11 is (-∞,-15.52) ∪ (-2.80,2.80) ∪ (15.52,∞).

Final Answer

latexTo solve the inequality, we make the assignments

Y_1=\operatorname{ABS}\left(0.14 x^2-13.72\right) \quad \text { and } \quad Y_2=\operatorname{ABS}(0.58 x)+11

and estimate the values of x for which the graph of $Y_1$ is above the graph of $Y_2$ (since we want $\mathrm{Y}_1$ greater than $\mathrm{Y}_2$ ). After perhaps several trials, we choose the viewing rectangle [-30,30 ,5] by [0, 40, 5], obtaining graphs similar to those in Figure 12. Since there is symmetry with respect to the y-axis, it is sufficient to find the x-coordinates of the points of intersection of the graphs for x>0. Using an intersect feature, we obtain $x \approx 2.80$ and $x \approx 15.52$ . Referring to Figure 12, we obtain the (approximate) solution
$(-\infty,-15.52) \cup(-2.80,2.80) \cup(15.52, \infty) .$