Let f(x)= _x^2^0t^2-8t+15e^tdt,x R.

Question

Let $f(x)={\int }_{{x}^{2}}^{0}\frac{{t}^{2}-8t+15}{{e}^{t}}dt,x\in R$. Then the numbers of local maximum and local minimum points of $f$, respectively, are :

Accepted Answer

We are given

$f(x)={\int }_{{x}^{2}}^{0}\frac{{t}^{2}-8t+15}{{e}^{t}}\, dt,\, x\in R.$

To find the local extrema, we first compute the derivative using the Fundamental Theorem of Calculus and the chain rule.

Rewrite the derivative: 1. Rewrite the derivative:

Let

$F(u)={\int }_{u}^{0}\frac{{t}^{2}-8t+15}{{e}^{t}}\, dt,$

so that

$f(x)=F({x}^{2}).$

Then by the chain rule,

${f}^{'}(x)={F}^{'}({x}^{2})\cdot 2x.$

Since

${F}^{'}(u)=\frac{{u}^{2}-8u+15}{{e}^{u}},$

we substitute $u={x}^{2}$ to get

${f}^{'}(x)=\frac{({x}^{2}{)}^{2}-8{x}^{2}+15}{{e}^{{x}^{2}}}\cdot 2x=\frac{2x\, ({x}^{4}-8{x}^{2}+15)}{{e}^{{x}^{2}}}.$

Factor and Identify Critical Points: 2. Factor and Identify Critical Points:

Factor the polynomial ${x}^{4}-8{x}^{2}+15$ by writing it in terms of ${x}^{2}$. Let $y={x}^{2}$, then

${y}^{2}-8y+15=(y-3)(y-5)$

so that

${x}^{4}-8{x}^{2}+15=({x}^{2}-3)({x}^{2}-5).$

Thus,

${f}^{'}(x)=\frac{2x\, ({x}^{2}-3)({x}^{2}-5)}{{e}^{{x}^{2}}}.$

Since ${e}^{{x}^{2}}>0$ for all $x$, the zeros of ${f}^{'}(x)$ are determined by

$2x\, ({x}^{2}-3)({x}^{2}-5)=0.$

That gives the critical points:

$2x=0\Rightarrow x=0$,

${x}^{2}-3=0\Rightarrow x=\pm \sqrt{3}$,

${x}^{2}-5=0\Rightarrow x=\pm \sqrt{5}$.

Analyzing the Sign of ${f}^{'}(x)$: 3. Analyzing the Sign of ${f}^{'}(x)$:

We need to determine the nature (maximum or minimum) by looking at the sign changes of ${f}^{'}(x)$ on the intervals determined by the critical points $x=-\sqrt{5}$, $x=-\sqrt{3}$, $x=0$, $x=\sqrt{3}$, and $x=\sqrt{5}$. Notice that the factor $2x\, ({x}^{2}-3)({x}^{2}-5)$ will dictate the sign.

Let's define:

$h(x)=x\, ({x}^{2}-3)({x}^{2}-5).$

Examine the sign of $h(x)$ in each interval:

For $x$x$ is negative.

${x}\^{2}>5$ so ${x}\^{2}-3>0$ and ${x}\^{2}-5>0$.

Product: negative $	imes$ positive $	imes$ positive = negative.

Thus, ${f}\^{'}(x)For $-\sqrt{5}$x$ is negative.

${x}\^{2}$ is between 3 and 5 so ${x}\^{2}-3>0$ while ${x}\^{2}-5Product: negative $	imes$ positive $	imes$ negative = positive.

Thus, ${f}^{'}(x)>0$.

For $-\sqrt{3}$x$ is negative.

${x}\^{2}Product: negative $	imes$ negative $	imes$ negative = negative.

Thus, ${f}^{'}(x)For $0$x$ is positive.

${x}^{2}Product: positive $	imes$ negative $	imes$ negative = positive.

Thus, ${f}\^{'}(x)>0$.

For $\sqrt{3}$x$ is positive.

${x}^{2}$ is between 3 and 5 so ${x}^{2}-3>0$ while ${x}^{2}-5Product: positive $	imes$ positive $	imes$ negative = negative.

Thus, ${f}\^{'}(x)For $x>\sqrt{5}$:

$x$ is positive.

${x}^{2}>5$ so both ${x}^{2}-3>0$ and ${x}^{2}-5>0$.

Product: positive $	imes$ positive $	imes$ positive = positive.

Thus, ${f}^{'}(x)>0$.

Classify the Critical Points: 4. Classify the Critical Points:

At $x=-\sqrt{5}$: ${f}^{'}(x)$ changes from negative to positive 	o  local minimum.

At $x=-\sqrt{3}$: ${f}^{'}(x)$ changes from positive to negative 	o  local maximum.

At $x=0$: ${f}^{'}(x)$ changes from negative to positive 	o  local minimum.

At $x=\sqrt{3}$: ${f}^{'}(x)$ changes from positive to negative 	o  local maximum.

At $x=\sqrt{5}$: ${f}^{'}(x)$ changes from negative to positive 	o  local minimum.

Conclusion: 5. Conclusion:

There are local maximum points at $x=-\sqrt{3}$ and $x=\sqrt{3}$ (2 points in total), and local minimum points at $x=-\sqrt{5}$, $x=0$, and $x=\sqrt{5}$ (3 points in total).

Thus, the numbers of local maximum and local minimum points of $f$ are 2 and 3, respectively.

The correct option is Option C.

Solutions · 1