Question

hw 10 - product and quotient rules section 2.5: problem 6 (1 point)
let $f(x)=(x^{3}-3)(3x^{2}+7)$. find an equation for the tangent line to the graph of $f$ at $x = 3$.
tangent line: $y=square$
if you need to round to at least 2 decimal places.

Explanation:

Step1: Find the derivative using product - rule

The product - rule states that if $y = u(x)v(x)$, then $y^\prime=u^\prime(x)v(x)+u(x)v^\prime(x)$. Let $u(x)=x^{3}-3$ and $v(x)=3x^{2}+7$. Then $u^\prime(x) = 3x^{2}$ and $v^\prime(x)=6x$. So $f^\prime(x)=3x^{2}(3x^{2}+7)+(x^{3}-3)\times6x=9x^{4}+21x^{2}+6x^{4}-18x = 15x^{4}+21x^{2}-18x$.

Step2: Evaluate $f(3)$ and $f^\prime(3)$

First, find $f(3)$: $f(3)=(3^{3}-3)(3\times3^{2}+7)=(27 - 3)(3\times9 + 7)=24\times(27 + 7)=24\times34 = 816$.
Next, find $f^\prime(3)$: $f^\prime(3)=15\times3^{4}+21\times3^{2}-18\times3=15\times81+21\times9 - 54=1215+189 - 54=1350$.

Step3: Use the point - slope form of a line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})$ is a point on the line and $m$ is the slope. Here, $x_{1}=3$, $y_{1}=816$, and $m = 1350$. So $y-816=1350(x - 3)$.
Expand to get $y-816=1350x-4050$.
Then $y = 1350x-4050 + 816=1350x-3234$.

Answer:

$y = 1350x-3234$