Question

hw 9 - the tangent line section 2.4: problem 4 (1 point) let $f(x)=8x^{4}-2x^{2}-4$. (a) find $f(x)$. $f(x)=square$ (b) find the slope of the line tangent to the graph of $f$ at $x = 8$. slope at $x = 8:square$ (c) find an equation of the line tangent to the graph of $f$ at $x = 8$. tangent line: $y=square$

Explanation:

Step1: Differentiate using power - rule

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$. For $f(x)=8x^{4}-2x^{2}-4$, we have $f^\prime(x)=8\times4x^{3}-2\times2x^{1}-0 = 32x^{3}-4x$.

Step2: Find the slope at $x = 8$

Substitute $x = 8$ into $f^\prime(x)$. So, $f^\prime(8)=32\times8^{3}-4\times8=32\times512 - 32=16384-32 = 16352$.

Step3: Find the point on the function at $x = 8$

First, find $f(8)$: $f(8)=8\times8^{4}-2\times8^{2}-4=8\times4096-2\times64 - 4=32768-128 - 4=32636$. The point on the function is $(8,32636)$.

Step4: Find the equation of the tangent line

Use the point - slope form of a line $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(8,32636)$ and $m = 16352$. So, $y-32636=16352(x - 8)$. Expand to get $y-32636=16352x-130816$. Then $y = 16352x-98180$.

Answer:

(a) $32x^{3}-4x$
(b) $16352$
(c) $y = 16352x-98180$