Question

find the equation of the tangent line to the graph of the given function at the given value of x.

f(x) = \sqrt{x^2 + 9}; x = 4

y = \square
(type an expression using x as the variable.)

Explanation:

Step1: Find f(4)

$f(4)=\sqrt{4^2 + 9}=\sqrt{16 + 9}=5$

Step2: Compute derivative f'(x)

Rewrite $f(x)=(x^2 + 9)^{1/2}$.
$f'(x)=\frac{1}{2}(x^2 + 9)^{-1/2} \cdot 2x=\frac{x}{\sqrt{x^2 + 9}}$

Step3: Find slope at x=4

$f'(4)=\frac{4}{\sqrt{4^2 + 9}}=\frac{4}{5}$

Step4: Write tangent line equation

Use point-slope form: $y - y_1 = m(x - x_1)$
$y - 5 = \frac{4}{5}(x - 4)$
Simplify: $y = \frac{4}{5}x - \frac{16}{5} + 5 = \frac{4}{5}x + \frac{9}{5}$

Answer:

$\frac{4}{5}x + \frac{9}{5}$