Question

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

Answer 1

Decreasing.

Explanation:

Use the sign, positive or negative, of the first derivative of a function to determine whether the function is increasing or decreasing:

• If `f'(pi/3)<0`, then `f` is decreasing at `x=pi/3`.
• If `f'(pi/3)>0`, then `f` is increasing at `x=pi/3`.

So, we first must find `f'`. Note that in order to differentiate the `e^xsinx` term we will have to use the product rule.

After applying the product rule and differentiating the initial `x` term to get `1`, we see that

`f'(x)=1-sinxd/dx(e^x)-e^xd/dx(sinx)`

Recalling that `d/dx(e^x)=e^x` and `d/dx(sinx)=cosx`, this gives

`f'(x)=1-e^xsinx-e^xcosx`

So, to determine whether the function is increasing or decreasing, we evaluate the derivative at `x=pi/3`:

`f'(pi/3)=1-e^(pi//3)sin(pi/3)-e^(pi//3)cos(pi/3)approx-2.8927`

Since this is negative, the function is decreasing at `pi/3`.

We can check a graph of `f` `(`note that `pi/3approx1.0472` `)`:

graph{x-e^xsinx [-9.78, 10.22, -6.75, 3.25]}

At `xapprox1.0472`, the function is decreasing.