Question

1. finding slope locations let f(x)=4√x - x. a. find all points on the graph of f at which the tangent line is horizontal. b. find all points on the graph of f at which the tangent line has slope -1/2.

Explanation:

Step1: Find the derivative of $f(x)$

First, rewrite $f(x)=4x^{\frac{1}{2}}-x$. Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $f^\prime(x)=\frac{4}{2}x^{-\frac{1}{2}}-1 = \frac{2}{\sqrt{x}}-1$.

Step2: Solve for part a (horizontal tangent)

A horizontal tangent line has a slope of 0. Set $f^\prime(x)=0$. So, $\frac{2}{\sqrt{x}}-1 = 0$. Add 1 to both sides: $\frac{2}{\sqrt{x}}=1$. Cross - multiply to get $\sqrt{x}=2$. Square both sides, and we find $x = 4$. When $x = 4$, $f(4)=4\sqrt{4}-4=4\times2 - 4=4$. The point is $(4,4)$.

Step3: Solve for part b (slope of $-\frac{1}{2}$)

Set $f^\prime(x)=-\frac{1}{2}$. So, $\frac{2}{\sqrt{x}}-1=-\frac{1}{2}$. Add 1 to both sides: $\frac{2}{\sqrt{x}}=1-\frac{1}{2}=\frac{1}{2}$. Cross - multiply to get $\sqrt{x}=4$. Square both sides, and we find $x = 16$. When $x = 16$, $f(16)=4\sqrt{16}-16=4\times4 - 16=0$. The point is $(16,0)$.

Answer:

a. $(4,4)$
b. $(16,0)$