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

- If #f'# changes from
**positive to negative** at #x=c#, then #f# has a local **maximum** at #c#
- 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.

- If #f'(x)# is
**positive** on interval #I#, then #f# is **increasing** on #I#
- 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]}