# F(X)=e^(4X)+e^(-x)a. Find the intervals on which f is increasing and decreasing. b. Find the local minimum value of f. c.  Find the interval on which f is concave up.

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Given `f(x)=e^(4x)+e^(-x)` :

(1) `f'(x)=4e^(4x)-e^(-x)`

(2) `f''(x)=16e^(4x)+e^(-x)`

(3) A function is increasing if the first derivative is positive, and decreasing if the first derivative is negative.

`4e^(4x)-e^(-x)>0`

`4e^(4x)>e^(-x)`

`ln[4e^(4x)]>ln[e^(-x)]`

`ln4+lne^(4x)>lne^(-x)`

`ln4+4x > -x`

`x>(-ln4)/5~~-.2772588722`

(4) So we find `f'(x)<0` on `(-oo,(-ln4)/5)` , `f'(x)=0` at `x=(-ln4)/5` , and `f'(x)>0` on `((-ln4)/5,oo)`

The function is decreasing on `(-oo,(-ln4)/5)` , and increasing on `((-ln4)/5,oo)`

(5) The function has local extrema only at the critical points; i.e where `f'(x)=0` or fails to exist.

Here the only critical point is at `x=(-ln4)/5` . Using the first derivative test, this point is a local minimum.

(6) In order to determine concavity, we look at the function's second derivative. If the 2nd derivative is positive, the function is concave up, if negative concave down.

We have `f''(x)=16e^4x+e^(-x)>0 forall x` , so the function is concave up everywhere.

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