Question

find the slope of the tangent line to the graph of the given function at the given value of x.
$y = -7x^{1/2} + x^{3/2}; x=16$
the slope of the tangent line to the graph is .
(simplify your answer.)

Explanation:

Step1: Find derivative of the function

Use power rule: $\frac{d}{dx}x^n = nx^{n-1}$
$\frac{dy}{dx} = -7\times\frac{1}{2}x^{\frac{1}{2}-1} + \frac{3}{2}x^{\frac{3}{2}-1} = -\frac{7}{2}x^{-\frac{1}{2}} + \frac{3}{2}x^{\frac{1}{2}}$

Step2: Substitute $x=16$ into derivative

First, calculate each term:
$x^{-\frac{1}{2}} = \frac{1}{\sqrt{x}} = \frac{1}{\sqrt{16}} = \frac{1}{4}$
$x^{\frac{1}{2}} = \sqrt{x} = \sqrt{16} = 4$
Substitute into derivative:
$\frac{dy}{dx}\bigg|_{x=16} = -\frac{7}{2}\times\frac{1}{4} + \frac{3}{2}\times4$

Step3: Compute the final value

Calculate each term:
$-\frac{7}{2}\times\frac{1}{4} = -\frac{7}{8}$
$\frac{3}{2}\times4 = 6 = \frac{48}{8}$
Sum the terms:
$-\frac{7}{8} + \frac{48}{8} = \frac{41}{8}$

Answer:

$\frac{41}{8}$