Question

Is `f(x)=(3-e^(2x))/x` increasing or decreasing at `x=-1`?

Answer 1

`f` decreases at `x=-1`.

Explanation:

We know that, `f'(a) <0 rArr f" is "darr" at "x=a`.

Now, `f(x)=(3-e^(2x))/x=(3-e^(2x))x^-1`

To find `f'(x)`, we use the Product and Chain Rule.

`f'(x)={3-e^(2x)}d/dx(x^-1)+x^-1{d/dx(3-e^(2x)}`

`={3-e^(2x)}(-1x^-2)+x^-1{0-e^(2x)d/dx(2x)}`

`=(e^(2x)-3)/x^2-(2e^(2x))/x`

`:. f'(x)=(e^(2x)-2xe^(2x)-3)/x^2={(1-2x)e^(2x)-3}/x^2`

`rArr f'(-1)={(1+2)e^-2-3}/(-1)^2=3(e^-2-1)`

Since, `2ltelt3, 4lte^2lt9, 1/4gt1/e^2gt1/9,` we have,

`f'(-1) <0.`

Therefore, `f` decreases at `x=-1`.

Enjoy Maths.!