# 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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### 1 Answer

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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