Question

let (f(x)=12sqrt{x}-2x)
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 (-\frac{1}{2})
the tangent line is horizontal at the point(s) (simplify your answer. type an ordered - pair. use a comma to separate answers as needed)

Explanation:

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

First, rewrite $f(x)=12x^{\frac{1}{2}}-2x$. Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $f^\prime(x)=12\times\frac{1}{2}x^{-\frac{1}{2}}-2 = \frac{6}{\sqrt{x}}-2$.

Step2: Solve for part a (horizontal tangent)

A horizontal tangent line has a slope of 0. Set $f^\prime(x) = 0$. So, $\frac{6}{\sqrt{x}}-2=0$. Add 2 to both sides: $\frac{6}{\sqrt{x}}=2$. Cross - multiply to get $2\sqrt{x}=6$, then $\sqrt{x}=3$. Square both sides to find $x = 9$. Substitute $x = 9$ into $f(x)$: $f(9)=12\sqrt{9}-2\times9=12\times3 - 18=36 - 18 = 18$.

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

Set $f^\prime(x)=-\frac{1}{2}$. So, $\frac{6}{\sqrt{x}}-2=-\frac{1}{2}$. Add 2 to both sides: $\frac{6}{\sqrt{x}}=2-\frac{1}{2}=\frac{3}{2}$. Cross - multiply: $3\sqrt{x}=12$, then $\sqrt{x}=4$. Square both sides to get $x = 16$. Substitute $x = 16$ into $f(x)$: $f(16)=12\sqrt{16}-2\times16=12\times4-32 = 48 - 32 = 16$.

Answer:

a. $(9,18)$
b. $(16,16)$