Question

assignment 4: problem 2 (1 point) let $f(x)=2x + 2sqrt{x}$. then: $f(5)=square$. use the previous answer to find the equation of the tangent line to the curve $y = 2x+2sqrt{x}$ at the point $(5,f(5))$. write your answer in the form $y = mx + b$ where $m$ is the slope and $b$ is the y - intercept. tangent line equation: $square$. note: you can earn partial credit on this problem.

Explanation:

Step1: Differentiate the function

The derivative of $f(x)=2x + 2\sqrt{x}=2x+2x^{\frac{1}{2}}$. Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $f'(x)=2 + 2\times\frac{1}{2}x^{-\frac{1}{2}}=2+\frac{1}{\sqrt{x}}$.

Step2: Find $f'(5)$

Substitute $x = 5$ into $f'(x)$. So $f'(5)=2+\frac{1}{\sqrt{5}}=\frac{2\sqrt{5}+1}{\sqrt{5}}$.

Step3: Find $f(5)$

$f(5)=2\times5+2\sqrt{5}=10 + 2\sqrt{5}$.

Step4: Use the point - slope form to find the tangent line equation

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(5,f(5))$ and $m = f'(5)$.
$y-(10 + 2\sqrt{5})=(2+\frac{1}{\sqrt{5}})(x - 5)$
$y-(10 + 2\sqrt{5})=(2+\frac{1}{\sqrt{5}})x-(10+\sqrt{5})$
$y=(2+\frac{1}{\sqrt{5}})x-(10+\sqrt{5})+(10 + 2\sqrt{5})$
$y=(2+\frac{1}{\sqrt{5}})x+\sqrt{5}$

Answer:

$f'(5)=\frac{2\sqrt{5}+1}{\sqrt{5}}$, Tangent line equation: $y=(2+\frac{1}{\sqrt{5}})x+\sqrt{5}$