This document outlines the steps to find the maximum value of a function: 1. Take the derivative of the function to find critical points where it is increasing, decreasing, or constant. 2. Check the critical points by taking the second derivative to determine if they are local maximums or minimums. 3. Plug the critical points that are maximums back into the original function to find the maximum value.

5.  Use first derivative test to classify profit
maximization…

6.  The derivative is related to the slope of a
function.

7. On an interval in which a function ‘f’ is
continuous and differentiable, a function is …
 Increasing
If f’(x) is positive on that interval, f’(x) > 0
 Decreasing
If f’(x) is negative on that interval, f’(x) < 0
 Constant
If f’(x) = 0 on that interval

8.  Find critical numbers.
These determine the boundaries of your
intervals.
 Pick a random x-value in each interval .
 Determine the sign of the derivative on that
interval.

9.  Find the maximum value…
y = 4x3 + 2x2 + 1.

10.  Step 1: Differentiate the function, using the power rule.
Constant terms disappear under differentiation.
d/dx (4x3 + 2x2 + 1)
= 12x2 + 4x
The result, 12x2 + 4x, is the gradient of the function. This has
two zeros, which can be found through factoring.
12x2 + 4x
= 4x(3x+1)
which equals zero when
4x=0 3x+1=0
x = 0 x = -1/3
Here, (x=0) & (x=-1/3) are critical points.

11.  Step 2: Check each turning point (at x = 0 and x = -
1/3)to find out whether it is a maximum or a
minimum. To do this, differentiate a second time
and substitute in the x value of each turning point.
d/dx (12x2 + 4x)
= 24x + 4
At x = 0
24x + 4 = 4, which is greater than zero. This is a
minimum.
At x = -1/3
24x + 4 = -4, which is less than zero. This is a
maximum.

12.  Step 3: Find the corresponding y-coordinates
for the x-value (maximum) you found in Step
2 by substituting back into the original
function.
At x = -1/3
y = 4x3 + 2x2 + 1
= -4/27 + 2/9 + 1
= 29/27
Therefore the function has a maximum
value at (-1/3, 29/27).