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f(x)=7+9x^2-6x^3Find the local maximum and minimum values of f using both the First and...

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sllysmith | Student, Undergraduate | eNoter

Posted April 8, 2012 at 12:57 AM via web

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f(x)=7+9x^2-6x^3

Find the local maximum and minimum values of f using both the First and Second Derivative Tests.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted April 8, 2012 at 8:53 AM (Answer #1)

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You need to find derivative of function and then you need to investigate the intervals where the derivative is positive and negative such that:

`f'(x) = 18x - 18x^2`

You need to solve the equation f'(x) = 0 such that:

`f'(x) = 0 =gt 18x - 18x^2 = 0`

You should factor out 18x such that:

`18x(1 - x) = 0`

You need to solve equations 18x = 0 and 1-x = 0 such that:

`18x = 0 =gt x = 0`

`1-x=0 =gt x =1`

You need to use the rule of signs such that:

You need to remember that between the roots 0 and 1 the values of f'(x) have the opposite sign to the sign of coefficient of the largest power (-18), hence `f'(x) gt 0` .

You need to remember that for values of x smaller than 0 and larger than 1 the values of f'(x) have like signs to the sign of coefficient of the largest power (-18), hence `f'(x)lt 0` .

Hence, the function decreases over `(-oo,0),`  then it increases over `(0,1)`  and then it decreases again over `(1,oo).`

Hence, the function reaches its minimum at x=0 and it reaches its maximum at x=1.

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thez | Student, Grade 11 | eNoter

Posted April 8, 2012 at 4:10 PM (Answer #2)

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I dont think you have answered the question.... Let me do it properly with the derivatives

First derivative f'(x) = 18x-18x^2

A turning point requires the gradient to be 0, therefore as the above post mentioned that these points are x = 0 and x = 1

Second derivative f''(x) = 18-36x (you differentiate f'(x))

for x = 0, f''(x) > 0, therefore it is a local minimum

for x = 1, f''(X) < 0, therefore it is a local maximum

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