8. what is the slope of the tangent line to the graph of $f(x)=(x^{2}+1)(2x^{3}-3x^{2}+1)$ at the point $(2,25)$? how fast is the function changing when $x = 2$?

Question

Accepted Answer

# Explanation:
## Step1: Apply the product rule
The product rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u = x^{2}+1$, so $u^\prime = 2x$. Let $v=2x^{3}-3x^{2}+1$, so $v^\prime=6x^{2}-6x$. Then $f^\prime(x)=(2x)(2x^{3}-3x^{2}+1)+(x^{2}+1)(6x^{2}-6x)$.
## Step2: Simplify $f^\prime(x)$
$$
\begin{align*}
f^\prime(x)&=4x^{4}-6x^{3}+2x + 6x^{4}-6x^{3}+6x^{2}-6x\
&=(4x^{4}+6x^{4})+(-6x^{3}-6x^{3})+6x^{2}+(2x - 6x)\
&=10x^{4}-12x^{3}+6x^{2}-4x
\end{align*}
$$
## Step3: Evaluate $f^\prime(x)$ at $x = 2$
Substitute $x = 2$ into $f^\prime(x)$:
$$
\begin{align*}
f^\prime(2)&=10\times(2)^{4}-12\times(2)^{3}+6\times(2)^{2}-4\times(2)\
&=10\times16-12\times8 + 6\times4-8\
&=160-96+24 - 8\
&=80
\end{align*}
$$

# Answer:
The slope of the tangent line at the point $(2,25)$ is $80$, and the function is changing at a rate of $80$ when $x = 2$.

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QUESTION IMAGE

Question

Explanation:

Step1: Apply the product rule

Step2: Simplify \(f^\prime(x)\)

Step3: Evaluate \(f^\prime(x)\) at \(x = 2\)

Answer: