`f(x) = e^(2x) + e^(-x)` (a) Find the intervals on which `f` is increasing or decreasing. (b) Find the local maximum and minimum values of `f`. (c) Find the intervals of concavity and the...

`f(x) = e^(2x) + e^(-x)` (a) Find the intervals on which `f` is increasing or decreasing. (b) Find the local maximum and minimum values of `f`. (c) Find the intervals of concavity and the inflection points.

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Textbook Question

Chapter 4, 4.3 - Problem 15 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to determine where the function increases or decreases, hence, you need to verify where f'(x)>0 or f'(x)<0.

You need to determine derivative of the function:

`f'(x) = 2e^(2x) - e^(-x)`

Putting f'(x) = 0, yields:

`2e^(2x) - e^(-x) = 0`

`2e^(2x) - 1/(e^x) = 0 => 2e^(3x) - 1 = 0 => e^(3x) = 1/2 => e^x = root(3)(1/2)`

`x = ln root(3)(1/2)`

Hence, the function decreases for `x in (-oo, ln root(3)(1/2))` and the function increases for `x in ( ln root(3)(1/2), oo)` .

b)You need to remember that the function reaches it's extrema for x, such as f'(x) = 0.

The function reaches it's minimum point at `x = ln root(3)(1/2).`

c) The function is concave up for f''(x)>0 and concave down for f''(x)<0.

Evaluating f''(x) yields:

`f''(x) = 4e^(2x) + e^(-x)`

`4e^(2x) + e^(-x) = 0=> 4e^(2x) + 1/(e^x) = 0 => 4e^(3x) + 1 = 0`

`e^x = root(3)(-1/4) = -root(3)(1/4)` impossible since `e^x>0` for all `x in R.`

Hence, the function has no inflection points and it is concave up for `x in R` , since `f''(x) = 4e^(2x) + e^(-x) > 0` for all `x in R` .

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