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…

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 $x = 2$?

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 $x = 2$?

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).