homework4: problem 8 (1 point) let f = 2x^2 + 4 and find the values below 1. f(x + h)= 2. (f(x + h)-f(x))=…

Explanation:

Step1: Find $f(x + h)$

Substitute $x+h$ into $f(x)$: [ \begin{align*} f(x + h)&=2(x + h)^2+4\ &=2(x^{2}+2xh+h^{2})+4\ &=2x^{2}+4xh + 2h^{2}+4 \end{align*} ]

Step2: Find $f(x + h)-f(x)$

[ \begin{align*} f(x + h)-f(x)&=(2x^{2}+4xh + 2h^{2}+4)-(2x^{2}+4)\ &=2x^{2}+4xh + 2h^{2}+4 - 2x^{2}-4\ &=4xh+2h^{2} \end{align*} ]

Step3: Find $\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$

[ \begin{align*} \lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}&=\lim_{h\rightarrow0}\frac{4xh + 2h^{2}}{h}\ &=\lim_{h\rightarrow0}(4x + 2h)\ &=4x \end{align*} ]

Step4: Find the equation of the tangent - line at $x = 5$

First, find the slope of the tangent - line at $x = 5$. Substitute $x = 5$ into the derivative. The derivative $f^\prime(x)=4x$, so $m=f^\prime(5)=20$. Next, find the $y$ - value when $x = 5$. $f(5)=2\times5^{2}+4=2\times25 + 4=50 + 4=54$. Use the point - slope form of a line $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(5,54)$ and $m = 20$. [ \begin{align*} y-54&=20(x - 5)\ y-54&=20x-100\ y&=20x - 46 \end{align*} ]

Answer:

1. $2x^{2}+4xh + 2h^{2}+4$
2. $4xh+2h^{2}$
3. $4x$
4. $y = 20x-46$