Solution assignment 03 Optimizing f(x)

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Assignment 3

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 sign of the first derivative.

Solution

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

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

thus:

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

and the solution of this quadratic equation is:

$x=2$ or $x=3$

Close to $x=2$ the derivative has the following signs + $0$ - and thus $x=2$ is a maximum. Close to $x=3$ the derivative has the following signs - $0$ + and thus this point is a minimum.

Return to Assignments Optimizing $f(x)$

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Solution assignment 03 Optimizing f(x)

Assignment 3

Solution

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