uation of the line tangent to the graph of f(x)=12 sin(7x)-56\sqrt3{x + 1} at x = 0. next item pts remaining.

Question

Accepted Answer

# Explanation:
## Step1: Find the derivative of $f(x)$.
Use the sum - difference rule and chain rule. The derivative of $\sin(u)$ is $\cos(u)\cdot u'$ and the derivative of $\sqrt[3]{x + 1}=(x + 1)^{\frac{1}{3}}$ is $\frac{1}{3}(x + 1)^{-\frac{2}{3}}$.
$$
\begin{align*}
f(x)&=12\sin(7x)-5\sqrt[3]{x + 1}\
f'(x)&=12\cos(7x)\cdot7-5\cdot\frac{1}{3}(x + 1)^{-\frac{2}{3}}\
f'(x)&=84\cos(7x)-\frac{5}{3(x + 1)^{\frac{2}{3}}}
\end{align*}
$$
## Step2: Evaluate $f'(x)$ at $x = 0$.
Substitute $x = 0$ into $f'(x)$.
$$
\begin{align*}
f'(0)&=84\cos(0)-\frac{5}{3(0 + 1)^{\frac{2}{3}}}\
&=84\times1-\frac{5}{3}\
&=\frac{252 - 5}{3}\
&=\frac{247}{3}
\end{align*}
$$
## Step3: Find $f(0)$.
$$
\begin{align*}
f(0)&=12\sin(0)-5\sqrt[3]{0+1}\
&=12\times0 - 5\times1\
&=- 5
\end{align*}
$$
## Step4: Use the point - slope form $y - y_0=m(x - x_0)$ for the tangent line.
Here $x_0 = 0$, $y_0=f(0)=-5$ and $m = f'(0)=\frac{247}{3}$.
$$
\begin{align*}
y-(-5)&=\frac{247}{3}(x - 0)\
y + 5&=\frac{247}{3}x\
y&=\frac{247}{3}x-5
\end{align*}
$$
# Answer:
$y=\frac{247}{3}x - 5$

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QUESTION IMAGE

Question

Explanation:

Step1: Find the derivative of \(f(x)\).

Step2: Evaluate \(f'(x)\) at \(x = 0\).

Step3: Find \(f(0)\).

Step4: Use the point - slope form \(y - y_0=m(x - x_0)\) for the tangent line.

Answer: