Question

find the intervals on which f is increasing or decreasing on the function f(x)= e^2x+e^-x

Answer 1

Have you computed the critical points?

Answer 2

no, i know i need to find f"(x) to find them but im not sure how

Answer 3

no you only need \(f'(x)\)

Answer 4

oh yeah f'(X), sorry typo

Answer 5

factor out an \(e^{-x}\) from your derivative

Answer 6

you should get
\[e^{-x}(2e^{3x}-1)\] and then it is easy to find the critical point, since \(e^{-x}\) is never zero, set
\[2e^{3x}-1=0\] and solve for \(x\)

Answer 7

if you just look at the derivative without factoring, it is hard to tell where it is zero

Answer 8

Find the domain in which f'(x) > 0. f(x) will be increasing in that domain.
Find the domain in which f'(x) <0. f(x) will be decreasing in that domain.
The critical point will be when f'(x) = 0 and that will separate the two domains.

Answer 9

where did the 2e^3x come from?

Answer 10

@satellite73

Answer 11

f(x) = e^(2x) + e^(-x)
f'(x) = 2e^(2x) - e^(-x) = 0
2e^(2x) = e^(-x)
multiply both sides by e^x
2e^(3x) = 1
e^(3x) = 1/2 = 0.5
Take ln on both sides
3x = ln(0.5)
x = ln(0.5) / 3

Answer 12

Ok, so if \(\large f(x) = e^{2x}+e^{-x}\), then by the chain rule, you should get \(\large f^{\prime}(x) = 2e^{2x}-e^{-x}\). Now, to find the critical points, we set the equation equal to zero and solve:
\[\large \begin{aligned}2e^{2x}-e^{-x} = 0 &\implies e^{-x}(2e^{3x}-1) = 0\\ &\implies 2e^{3x}-1 = 0\\ & \implies e^{3x}=\frac{1}{2} \\ &\implies 3x=\ln(1/2) = -\ln 2\\ &\implies x=-\frac{\ln 2}{3}\end{aligned}\].
Now what you want to do is fill in the following table:
\[\begin{array}{c|c|c}\text{interval} & \left(-\infty,-\frac{\ln 2}{3}\right) & \left(\frac{\ln 2}{3},\infty\right)\\\hline \text{sign of $f^{\prime}(x)$} & & \\\hline \text{behavior of $f(x)$} \end{array}\] For the row called "sign of \(f^{\prime}(x)\)", what you do is take an x value in the respective interval and plug it into the derivative. If the resulting value is positive, put a + there, and if it's negative, but a - there. Wherever \(\large f^{\prime}(x) >0\), \(\large f(x)\) is increasing there and wherever \(\large f^{\prime}(x) <0\), \(\large f(x)\) is decreasing there.

I hope this makes sense! :-)

Answer 13

That second interval in the table should read \(\large \left(-\dfrac{\ln 2}{3},\infty\right)\); sorry about that! :-/

Answer 14

Wow this makes perfect sense thank you ^-^