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