You need to use derivative of the function to check its monotony such that:

`f'(x)>0=> f(x) ` increases

`f'(x)<0 => f(x) ` decreases

Hence, you need to find derivative of the given function, using the chain rule, such that:

`f'(x) = (e^(5x) + e^(−x))' => f'(x) = e^(5x)*(5x)' + e^(-x)*(-x)'`

`f'(x)...

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You need to use derivative of the function to check its monotony such that:

`f'(x)>0=> f(x) ` increases

`f'(x)<0 => f(x) ` decreases

Hence, you need to find derivative of the given function, using the chain rule, such that:

`f'(x) = (e^(5x) + e^(−x))' => f'(x) = e^(5x)*(5x)' + e^(-x)*(-x)'`

`f'(x) = 5e^(5x) - e^(-x)`

You need to use negative power property to write `e^(-x)` such that:

`e^(-x) = 1/(e^x)`

`f'(x) = 5e^(5x) - 1/e^x => f'(x) = (5e^(5x)*e^x - 1)/e^x`

You need to use the following property of exponentials such that:

`a^b*a^c = a^(b+c)`

`f'(x) = (5e^(5x+x) - 1)/e^x => f'(x) = (5e^(6x) - 1)/e^x `

You need to solve for x the equation `f'(x) = 0 ` such that:

`(5e^(6x) - 1)/e^x = 0 => 5e^(6x) - 1 = 0 => 5e^(6x) = 1`

`e^(6x) = 1/5`

Taking logarithms both sides yields:

`ln e^(6x) = ln (1/5) => 6x ln e = ln (1/5) => x = (ln (1/5))/6`

The graph shows to you that the derivative is negative to the left of its root `x = (ln (1/5))/6` and it is positive to the right.

**Hence, the function increases if `x in ((ln (1/5))/6, oo)` and it decreases if `x in (-oo, (ln (1/5))/6).` **