Question

1. finding slope locations let f(x) = 2e^x - 6x. 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 12.

Explanation:

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

The derivative of $y = 2e^{x}-6x$ using the rules $\frac{d}{dx}(e^{x})=e^{x}$ and $\frac{d}{dx}(ax)=a$ is $f^\prime(x)=2e^{x}-6$.

Step2: Solve for part a (horizontal tangent)

A horizontal tangent has a slope of 0. Set $f^\prime(x) = 0$.
$$2e^{x}-6 = 0$$
Add 6 to both sides: $2e^{x}=6$. Then divide by 2: $e^{x}=3$. Take the natural - logarithm of both sides: $x=\ln(3)$.
Substitute $x = \ln(3)$ into $f(x)$: $f(\ln(3))=2e^{\ln(3)}-6\ln(3)=2\times3 - 6\ln(3)=6 - 6\ln(3)$. So the point is $(\ln(3),6 - 6\ln(3))$.

Step3: Solve for part b (slope = 12)

Set $f^\prime(x)=12$.
$$2e^{x}-6 = 12$$
Add 6 to both sides: $2e^{x}=18$. Divide by 2: $e^{x}=9$. Take the natural - logarithm of both sides: $x=\ln(9)$.
Substitute $x=\ln(9)$ into $f(x)$: $f(\ln(9))=2e^{\ln(9)}-6\ln(9)=2\times9-6\ln(9)=18 - 6\ln(9)$. So the point is $(\ln(9),18 - 6\ln(9))$.

Answer:

a. $(\ln(3),6 - 6\ln(3))$
b. $(\ln(9),18 - 6\ln(9))$