Solution assignment 04 Optimizing f(x)

Assignment 4

Determine the extreme of the function:

$y=2x^3-15x^2+36x-2$

and verify whether it is a maximum or a minimum by using the second derivative.

Solution

In order to find the candidates of extremes we have to calculate the derivative of the function and solve the following equation:

$y'=6x^2-30x+36=0$

thus:

$x^2-5x+6=0$

and this equation has the solution:

$x=2$ or $x=3$

The second derivative is:

$y''=2x-5$

For $x=2$ the second derivative is negative and thus there is a maximum for $x=2$ .
For $x=3$ the second derivative is positive and thus there is a minimum for $x=3$ .

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Solution assignment 04 Optimizing f(x)

Assignment 4

Solution

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