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

Dec 19, 2016

Use the First Derivative Test.

#### Explanation:

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 $f ' \left(x\right)$ is positive on interval $I$, then $f$ is increasing on $I$
2. If $f ' \left(x\right)$ is negative on interval $I$, then $f$ is decreasing on $I$

The function we are given is $f \left(x\right) = 2 {x}^{3} - 5 {x}^{2} - 4 x + 7$

Taking the first derivative,

$f ' \left(x\right) = 6 {x}^{2} - 10 x - 4$

$\implies 0 = 6 {x}^{2} - 10 x - 4$

We can factor:

$0 = \left(3 x + 1\right) \left(2 x - 4\right)$

And solving for $x$,

$x = - \frac{1}{3} , x = 2$

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 ${x}_{1} = - \frac{1}{3}$ and ${x}_{2} = 2$

We now check $f ' \left(x\right)$ at $x = - 1 , 0 , 3$

$f ' \left(- 1\right)$ yields a positive answer.
$f ' \left(0\right)$ yields a negative answer.
$f ' \left(3\right)$ 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 $x = - \frac{1}{3}$ 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.

$f \left(- \frac{1}{3}\right) = \frac{208}{27}$
$f \left(2\right) = - 5$

Thus, we have a local maximum at $\left(- \frac{1}{3} , \frac{208}{27}\right)$ and a local minimum at $\left(2 , - 5\right)$.

We can see this is correct if we look at the graph of $f \left(x\right)$:

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