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![EXAMPLE Find the extreme values of f(x)= 2x3 − 15x2 + 24x + 7 on [0, 6].
Solution The extreme values occur at critical points or endpoints, so we can break up the problem neatly into two steps.
Step 1. Find the critical points.
The function f(x) is differentiable, so we find the critical points by solving
The critical points are c = 1 and 4.
Step 2. Calculate the y-value at the critical points and endpoints and compare.
The maximum of f(x) on [0, 6] is the largest of the values in this table, namely f(6) = 43
(Figure 9). Similarly, the minimum is f(4) = − 9.](https://image.slidesharecdn.com/extremevalues-110608183917-phpapp02/75/Extreme-values-1-2048.jpg)
The document provides a solution to find the extreme values of the function f(x)= 2x3 − 15x2 + 24x + 7 on the interval [0, 6]. It outlines a two step process of first finding the critical points by solving the derivative, which are 1 and 4. The second step is to calculate the y-values at these critical points and the endpoints and compare them, with the maximum being f(6)=43 and the minimum being f(4)=-9.
![EXAMPLE Find the extreme values of f(x)= 2x3 − 15x2 + 24x + 7 on [0, 6].
Solution The extreme values occur at critical points or endpoints, so we can break up the problem neatly into two steps.
Step 1. Find the critical points.
The function f(x) is differentiable, so we find the critical points by solving
The critical points are c = 1 and 4.
Step 2. Calculate the y-value at the critical points and endpoints and compare.
The maximum of f(x) on [0, 6] is the largest of the values in this table, namely f(6) = 43
(Figure 9). Similarly, the minimum is f(4) = − 9.](https://image.slidesharecdn.com/extremevalues-110608183917-phpapp02/75/Extreme-values-1-2048.jpg)