Let f(x)=_0^x^2 t^2-8 t+15e^t dt, x R.

Question

Let $f(x)=\int_0^{x^2} \frac{\mathrm{t}^2-8 \mathrm{t}+15}{\mathrm{e}^{\mathrm{t}}} \mathrm{dt}, x \in \mathbf{R}$. Then the numbers of local maximum and local minimum points of $f$, respectively, are :

Accepted Answer

We are given

$$ f(x)=\int_0^{x^2} \frac{t^2-8t+15}{e^t}\,dt, \quad x\in\mathbb{R}. $$

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

$$ F(u)=\int_0^u \frac{t^2-8t+15}{e^t}\,dt, $$

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

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

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

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

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

so that

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

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 $\times$ positive $\times$ 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\^2-5

Product: negative $\times$ positive $\times$ negative = positive.

Thus, $f'(x)>0$.

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

$x\^2

Product: negative $\times$ negative $\times$ negative = negative.

Thus, $f'(x)

For $0 $x$ is positive.

$x^2

Product: positive $\times$ negative $\times$ 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^2-5

Product: positive $\times$ positive $\times$ 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 $\times$ positive $\times$ 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 \to  local minimum.

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

At $x=0$: $f'(x)$ changes from negative to positive \to  local minimum.

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

At $x=\sqrt{5}$: $f'(x)$ changes from negative to positive \to  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