# Calculus Consider the equation below. f(x) = e^(3x) + e^(−x)   (a) Find the intervals on which f is increasing. (Enter your answer using interval notation.) 1 Find the interval on which f is decreasing. (Enter your answer using interval notation.) 2 (b) Find the local minimum value of f. 3 (c) Find the interval on which f is concave up. (Enter your answer using interval notation.) 4 `f(x)=e^(3x)+e^(-x) => f'(x)=3e^(3x)-e^(-x) =>f''(x)=9e^(3x)+e^(-x)`

f(x) is increasing if f'(x)>0, and decreasing if f'(x)<0.

`f'(x)<0=>3e^(3x)-e^(-x)<0=>`

`3e^(3x)<e^(-x)=>3e^(3x)<1/e^x=>`

`3e^(3x)*e^x<1=>3e^(4x)<1=>`

`e^4x<1/3=>4x<ln(1/3)=>x<1/4*ln(1/3)=>x<1/4*ln(3^(-1))=>`

`x<[-ln3]/4`

Hence

f(x) is decreasing over `(-oo,[-ln3]/4)`

f(x) is increasing over `([-ln3]/4,+oo)`

local minimum is at `[-ln3]/4`

c) We already found that

`f''(x)=9e^(3x)+e^(-x)`

That function is positive for all values of x, hence...

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`f(x)=e^(3x)+e^(-x) => f'(x)=3e^(3x)-e^(-x) =>f''(x)=9e^(3x)+e^(-x)`

f(x) is increasing if f'(x)>0, and decreasing if f'(x)<0.

`f'(x)<0=>3e^(3x)-e^(-x)<0=>`

`3e^(3x)<e^(-x)=>3e^(3x)<1/e^x=>`

`3e^(3x)*e^x<1=>3e^(4x)<1=>`

`e^4x<1/3=>4x<ln(1/3)=>x<1/4*ln(1/3)=>x<1/4*ln(3^(-1))=>`

`x<[-ln3]/4`

Hence

f(x) is decreasing over `(-oo,[-ln3]/4)`

f(x) is increasing over `([-ln3]/4,+oo)`

local minimum is at `[-ln3]/4`

c) We already found that

`f''(x)=9e^(3x)+e^(-x)`

That function is positive for all values of x, hence the function is concave up over `(-oo,+oo)`

The following graph confirms our findings

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