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Q. 26

Expert-verifiedFound in: Page 171

Book edition
6th

Author(s)
Sullivan

Pages
1200 pages

ISBN
9780321795465

In Problems 25–32, use the given functions f and g.

$\left(a\right)f\left(x\right)=0\phantom{\rule{0ex}{0ex}}\left(b\right)g\left(x\right)=0\phantom{\rule{0ex}{0ex}}\left(c\right)f\left(x\right)=g\left(x\right)\phantom{\rule{0ex}{0ex}}\left(d\right)f\left(x\right)>0$

$\left(e\right)g\left(x\right)\le 0\phantom{\rule{0ex}{0ex}}\left(f\right)f\left(x\right)>g\left(x\right)\phantom{\rule{0ex}{0ex}}\left(g\right)f\left(x\right)\ge 1$

$f\left(x\right)=-{x}^{2}+3\phantom{\rule{0ex}{0ex}}g\left(x\right)=-3x+3$

The required solution sets are

(a) $x=-\sqrt{3},\sqrt{3}$

(b) $x=1$

(c) $x=0,3$

(d) $\left\{x:-\sqrt{3}<x<\sqrt{3}\right\}$

(e) $x\ge 1$

(f) $\left\{x:0<x<3\right\}$

(g) $\left\{x:-\sqrt{2}\le x\le \sqrt{2}\right\}$

- The given functions are $f\left(x\right)=-{x}^{2}+3\phantom{\rule{0ex}{0ex}}g\left(x\right)=-3x+3$.
- The equation is $f\left(x\right)=-{x}^{2}+3\phantom{\rule{0ex}{0ex}}g\left(x\right)=-3x+3$.

- Plot the graph of the function.

- From the graph, it can be observed that $f\left(x\right)=0$ when $x=-\sqrt{3},\sqrt{3}$.

- The given functions are $f\left(x\right)=-{x}^{2}+3\phantom{\rule{0ex}{0ex}}g\left(x\right)=-3x+3$
- The equation is $g\left(x\right)=0$.

- Plot the line in the graph obtained for the first function.

- From the graph, it can be observed that $g\left(x\right)=0$when $x=1$.

- The given functions are $f\left(x\right)=-{x}^{2}+3\phantom{\rule{0ex}{0ex}}g\left(x\right)=-3x+3$
- The equation is $f\left(x\right)=g\left(x\right)$.

- For $f\left(x\right)=g\left(x\right)$, the curves of both the functions must intersect.
- From the graph, it can be observed that the functions intersect at $(0,3)and(3,-6)$.
- So, $f\left(x\right)=g\left(x\right)$at $x=0,3$.

- The given functions are $f\left(x\right)=-{x}^{2}+3\phantom{\rule{0ex}{0ex}}g\left(x\right)=-3x+3$
- The inequality is $f\left(x\right)>0$.

- The inequality holds when the curve of the function is above the horizontal axis.
- According to the graph obtained in step 2 of part (b), the curve is above the horizontal axis when $-\sqrt{3}<x<\sqrt{3}$.
- So, the solution set is $\left\{x:-\sqrt{3}<x<\sqrt{3}\right\}$.

- The given functions are $f\left(x\right)=-{x}^{2}+3\phantom{\rule{0ex}{0ex}}g\left(x\right)=-3x+3$
- The inequality is $g\left(x\right)\le 0$.

- $g\left(x\right)\le 0$ when the line of the function is on or below the horizontal axis.
- From the graph, the line is on or below the axis for $x\ge 1$.

- The given functions are $f\left(x\right)=-{x}^{2}+3\phantom{\rule{0ex}{0ex}}g\left(x\right)=-3x+3$
- The inequality is $f\left(x\right)>g\left(x\right)$.

- The inequality holds when the curve lies above the line on the graph.
- From the graph, it can be observed that the curve is above the line when $0<x<3$.
- So, the solution set of the inequality is $\left\{x:0<x<3\right\}$.

- The given functions are $f\left(x\right)=-{x}^{2}+3\phantom{\rule{0ex}{0ex}}g\left(x\right)=-3x+3$
- The inequality is $f\left(x\right)\ge 1$.

- The inequality holds when the curve lies above the value 1 on the vertical axis.
- From the graph, the curve is above 1 when $-\sqrt{2}\le x\le \sqrt{2}$.
- So, the solution set of the inequality is $\left\{x:-\sqrt{2}\le x\le \sqrt{2}\right\}$.

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