For the function $$ f(x)=3 e^{-x^{3}} $$, find $$ f^{\prime \prime}(x) $$. Then find $$ f^{\prime \prime}(0) $$ and $$ f^{\prime \prime}(2) $$. $$ f^{\prime \prime}(x)=\square $$ Select the correct choice below and fill in any answer boxes in your choice. A. $$ f^{\prime \prime}(0)=\square $$ (Simplify your answer.) B. $$ f^{\prime \prime}(0) $$ is undefined. Select the correct choice below and fill in any answer boxes in your choice. A. $$ f^{\prime \prime}(2)=\square $$ $$ \quad(T y p e ~ a n ~ i n t e g e r ~ o r ~ a ~ d e c i m a l . ~ D o ~ n o t ~ r o u n d ~ u n t i l ~ t h e ~ f i n a l ~ a n s w e r . ~ T h e n ~ r o u n d ~ t o ~ f o u r ~ d e c i m a l ~ p l a c e s ~ a s ~ n e e d e d .) $$. B. $$ f^{\prime \prime}(2) $$ is undefined.

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UpStudy AI · Accepted Answer

We start with
$$
f(x)=3e^{-x^3}.
$$

**Step 1. Compute $$ f'(x) $$:**

Differentiate using the chain rule:
$$
f'(x)=\frac{d}{dx}\left[3e^{-x^3}\right]=3\cdot e^{-x^3}\cdot (-3x^2)=-9x^2 e^{-x^3}.
$$

**Step 2. Compute $$ f''(x) $$:**

Differentiate $$ f'(x)=-9x^2 e^{-x^3} $$ using the product rule:
$$
u(x)=-9x^2 \quad \text{and} \quad v(x)=e^{-x^3}.
$$
Their derivatives are:
$$
u'(x)=-18x \quad \text{and} \quad v'(x)=e^{-x^3}\cdot (-3x^2)=-3x^2 e^{-x^3}.
$$
Thus,
$$
\begin{aligned}
f''(x) &= u'(x)v(x) + u(x)v'(x) \
       &= (-18x)e^{-x^3} + (-9x^2)(-3x^2 e^{-x^3}) \
       &= -18x e^{-x^3} + 27x^4 e^{-x^3} \
       &= e^{-x^3}\left(-18x+27x^4\right) \
       &= 9x e^{-x^3}\left( -2+3x^3 \right).
\end{aligned}
$$
Thus, 
$$
f''(x)=9x e^{-x^3}(3x^3-2).
$$

**Step 3. Evaluate $$ f''(0) $$:**

Substitute $$ x=0 $$:
$$
f''(0)=9\cdot 0\cdot e^{0}\cdot (3\cdot 0^3-2)=0.
$$

So, the answer to the first selection is:

A. $$ f^{\prime \prime}(0)=0 $$

**Step 4. Evaluate $$ f''(2) $$:**

Substitute $$ x=2 $$:
$$
\begin{aligned}
f''(2) &=9\cdot 2\cdot e^{-(2)^3}\left(3(2)^3-2\right) \
       &=18e^{-8}\left(3\cdot 8-2\right) \
       &=18e^{-8}(24-2) \
       &=18e^{-8}\cdot 22 \
       &=396e^{-8}.
\end{aligned}
$$

Numerically, $$ e^{-8}\approx 0.00033546 $$, so
$$
f''(2)\approx396\cdot 0.00033546\approx0.1328.
$$

Thus, the answer to the second selection is:

A. $$ f^{\prime \prime}(2)=0.1328 $$
$$ f''(x) = 9x e^{-x^3}(3x^3 - 2) $$ $$ f''(0) = 0 $$ $$ f''(2) \approx 0.1328 $$

For the function , find . Then find and .

Select the correct choice below and fill in any answer boxes in your choice.
A. (Simplify your answer.)
B. is undefined.
Select the correct choice below and fill in any answer boxes in your choice.
A.
.
B. is undefined.

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For the function , find . Then find and . Select the correct choice below and fill in any answer boxes in your choice. A. (Simplify your answer.) B. is undefined. Select the correct choice below and fill in any answer boxes in your choice. A. . B. is undefined.