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. $$ f^{\prime \prime}(2)=\square $$ (Type an integer or a decimal. Do not round until the final answer. Then round to four decimal places as needed..) B. $$ f^{\prime \prime}(2) $$ is undefined.

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UpStudy AI · Accepted Answer

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

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

Using the chain rule,
$$
f'(x)=3e^{-x^3}\cdot \frac{d}{dx}(-x^3).
$$
Since
$$
\frac{d}{dx}(-x^3)=-3x^2,
$$
we have
$$
f'(x)=3e^{-x^3}\cdot (-3x^2)=-9x^2e^{-x^3}.
$$

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

Differentiate
$$
f'(x)=-9x^2e^{-x^3}
$$
using the product rule. Let
$$
u(x)=-9x^2 \quad \text{and} \quad v(x)=e^{-x^3}.
$$
Then
$$
u'(x)=-18x,
$$
and using the chain rule for $$ v(x) $$,
$$
v'(x)=e^{-x^3}\cdot (-3x^2)=-3x^2e^{-x^3}.
$$

Now the product rule gives:
$$
f''(x)=u'(x)v(x)+u(x)v'(x)=-18xe^{-x^3}+(-9x^2)(-3x^2e^{-x^3}).
$$
Simplify the second term:
$$
-9x^2\cdot (-3x^2e^{-x^3})=27x^4e^{-x^3}.
$$
Thus,
$$
f''(x)=-18xe^{-x^3}+27x^4e^{-x^3}.
$$

Factor out the common factor $$ e^{-x^3} $$:
$$
f''(x)=e^{-x^3}(27x^4-18x).
$$

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

Substitute $$ x=0 $$ into $$ f''(x) $$:
$$
f''(0)=e^{-0}(27\cdot0^4-18\cdot0)=1\cdot(0-0)=0.
$$

Thus, the correct choice for $$ f^{\prime \prime}(0) $$ is:

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

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

Substitute $$ x=2 $$ into $$ f''(x) $$:
$$
f''(2)=e^{-(2)^3}\left(27\cdot(2)^4-18\cdot2\right)=e^{-8}\left(27\cdot16-36\right).
$$
Calculate:
$$
27\cdot16=432,\quad 432-36=396.
$$
Thus,
$$
f''(2)=396e^{-8}.
$$

If needed as a decimal (rounded to four decimal places), we note that:
$$
e^{8}\approx2980.957987, \quad e^{-8}\approx\frac{1}{2980.957987}\approx0.00033546,
$$
and then,
$$
396e^{-8}\approx396\cdot0.00033546\approx0.1329.
$$

Thus, the correct choice for $$ f^{\prime \prime}(2) $$ is:

A. $$ f^{\prime \prime}(2)=396e^{-8} $$ (or approximately $$ 0.1329 $$ when rounded to four decimal places)

**Final Answers:**

$$
\boxed{f^{\prime \prime}(x)=e^{-x^3}(27x^4-18x),\quad f^{\prime \prime}(0)=0,\quad f^{\prime \prime}(2)=396e^{-8}.}
$$
$$ f^{\prime \prime}(x) = e^{-x^3}(27x^4 - 18x) $$ $$ f^{\prime \prime}(0) = 0 $$ $$ f^{\prime \prime}(2) = 396e^{-8} $$

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