$ f(x) = x^3 - 6x^2 - 15x + 8 $ for the interval $ [-2,6] $
My answer was :
$$ f'(x) = 3(x^2-4x-5) $$ $$ 3(x-5)(x+1) $$
so the critical points are $ x =5\; and \;x=-1 $
plugging them back with the end points into the original function
$$f(5) = 5^2 - 6(5)^2 - 15(5) +8 = -92 $$
$$ f(-1) = 16 $$
$$ f(-2) =6 $$
$$ f(6) = -82 $$
so i got the min is $-92$ and max is $16$ but the answer in the book says the min is $5$ and max is $-1$
$\endgroup$ 61 Answer
$\begingroup$Hint: You are almost right! Look at
The global maxima is different than the maximum value of the function $f(x)$. It is the value of $x$ which maximizes $f(x)$, not the maximum value itself.
$\endgroup$ 3
