Question

attempt 1: 10 attempts remaining. let (f(x)=x^{7}-3x^{5}+7x^{3}-3x - 3). (f(x)=) (f(2)=) (f(x)=) (f(2)=)

Explanation:

Step1: Apply power - rule for differentiation

The power - rule states that if $y = x^n$, then $y^\prime=nx^{n - 1}$. For $f(x)=x^{7}-3x^{5}+7x^{3}-3x - 3$, we have:
$f^\prime(x)=\frac{d}{dx}(x^{7})-3\frac{d}{dx}(x^{5})+7\frac{d}{dx}(x^{3})-3\frac{d}{dx}(x)-\frac{d}{dx}(3)$.
$f^\prime(x)=7x^{6}-15x^{4}+21x^{2}-3$.

Step2: Evaluate $f^\prime(2)$

Substitute $x = 2$ into $f^\prime(x)$:
$f^\prime(2)=7\times2^{6}-15\times2^{4}+21\times2^{2}-3$.
$=7\times64 - 15\times16+21\times4-3$.
$=448-240 + 84-3$.
$=289$.

Step3: Differentiate $f^\prime(x)$ to get $f^{\prime\prime}(x)$

Differentiate $f^\prime(x)=7x^{6}-15x^{4}+21x^{2}-3$ using the power - rule again.
$f^{\prime\prime}(x)=\frac{d}{dx}(7x^{6})-15\frac{d}{dx}(x^{4})+21\frac{d}{dx}(x^{2})-\frac{d}{dx}(3)$.
$f^{\prime\prime}(x)=42x^{5}-60x^{3}+42x$.

Step4: Evaluate $f^{\prime\prime}(2)$

Substitute $x = 2$ into $f^{\prime\prime}(x)$:
$f^{\prime\prime}(2)=42\times2^{5}-60\times2^{3}+42\times2$.
$=42\times32-60\times8 + 84$.
$=1344-480+84$.
$=948$.

Answer:

$f^\prime(x)=7x^{6}-15x^{4}+21x^{2}-3$
$f^\prime(2)=289$
$f^{\prime\prime}(x)=42x^{5}-60x^{3}+42x$
$f^{\prime\prime}(2)=948$