Question

let f(x)=6√x - 3x. 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 - 5/2. 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 $f(x)$

Given $f(x)=6\sqrt{x}-3x = 6x^{\frac{1}{2}}-3x$. Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $f^\prime(x)=6\times\frac{1}{2}x^{-\frac{1}{2}}-3=\frac{3}{\sqrt{x}}-3$.

Step2: Solve for when the tangent is horizontal in part a

A horizontal tangent line has a slope of $0$. Set $f^\prime(x) = 0$. So, $\frac{3}{\sqrt{x}}-3 = 0$. Add $3$ to both sides: $\frac{3}{\sqrt{x}}=3$. Cross - multiply to get $3 = 3\sqrt{x}$. Then $\sqrt{x}=1$, and $x = 1$. Substitute $x = 1$ into $f(x)$: $f(1)=6\sqrt{1}-3\times1=6 - 3=3$. The point is $(1,3)$.

Step3: Solve for when the slope is $-\frac{5}{2}$ in part b

Set $f^\prime(x)=-\frac{5}{2}$. So, $\frac{3}{\sqrt{x}}-3=-\frac{5}{2}$. Add $3$ to both sides: $\frac{3}{\sqrt{x}}=3-\frac{5}{2}=\frac{6 - 5}{2}=\frac{1}{2}$. Cross - multiply: $3\times2=\sqrt{x}$, so $\sqrt{x}=6$ and $x = 36$. Substitute $x = 36$ into $f(x)$: $f(36)=6\sqrt{36}-3\times36=6\times6-108=36 - 108=-72$. The point is $(36,-72)$.

Answer:

a. $(1,3)$
b. $(36,-72)$