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