How do you find the local max and min for `y=2x^3 - 5x^2 - 4x + 7`?

By the First Derivative Test , if some function `f` has a critical number at `c`, then:

1. If `f'` changes from positive to negative at `x=c`, then `f` has a local maximum at `c`
2. If `f'` changes from negative to positive at `x=c`, then `f` has a local minimum at `c`

To determine local extrema using the First Derivative Test, take the first derivative of the function and set it equal to `0`. Solving for all possible values of `x` which satisfy this will yield critical numbers. You can then test those critical numbers to determine if they represent a local maximum, minimum, or neither by checking the intervals of increase and decrease of the derivative relative to those points.

1. If `f'(x)` is positive on interval `I`, then `f` is increasing on `I`
2. If `f'(x)` is negative on interval `I`, then `f` is decreasing on `I`

The function we are given is `f(x)=2x^3-5x^2-4x+7`

Taking the first derivative,

`f'(x)=6x^2-10x-4`

`=>0=6x^2-10x-4`

We can factor:

`0=(3x+1)(2x-4)`

And solving for `x`,

`x=-1/3, x=2`

These are our critical numbers. We now pick test points near these values for `x` (I would recommend integers) and check to see if they yield a positive or negative first derivative. Note that we want test points which fall on either side of our critical numbers. I would use test points of `-1,0,3`. I good way to organize this is with a number line.

Where `x_1=-1/3` and `x_2=2`

We now check `f'(x)` at `x=-1,0,3`

`f'(-1)` yields a positive answer.
`f'(0)` yields a negative answer.
`f'(3)` yields a positive answer.

Thus, `f` is increasing at `x=3` and `x=-1`, but decreasing at `x=0`. We can draw lines with positive or negative slopes respective to the signs of the derivatives to help us visualize where we might have a maximum or minimum (this is not necessary by any means, but can be a useful strategy until you are comfortable finding extrema).

We can see `x_1` appears to be at the top of peak of the lines on either side. We know that `x=-1/3` is a local maximum. Similarly, we can see that at `x_2` appears to be at the bottom of a peak of the lines on either side. We know that `x=2` is a local minimum.

We can then put those values of `x` back into the original function to determine the corresponding `y` values and get coordinates.

`f(-1/3)=208/27`
`f(2)=-5`

Thus, we have a local maximum at `(-1/3,208/27)` and a local minimum at `(2,-5)`.

We can see this is correct if we look at the graph of `f(x)`:

graph{2x^3-5x^2-4x+7 [-10, 10, -5, 5]}