Find the Max/Min values:
$f(x)=x^3-x^2-8x+1$ on $[-2,2]$
$$\begin{align}f'(x)&=3x^2-2x-8\\&=3x^2+4x-6x-8\\&=x(3x+4)-2(3x+4)\\&=(x-2)(3x+4)\end{align}$$
Thus, $x=-2,0.75,2$
Now sub these into the original function
$f(2)=2^3-2^2-8(2)+1=-11$
$f(0.75)=0.75^3-0.75^2-8(0.75)+1=-5.14$
$f(-2)=(-2)^3-(-2)^2-8(-2)+1=5$
Is the minimum $-11$ and the maximum $5$?
Any idea as to how I would go about $f(x)=x^5+x+1$ on $[-1,1]$?
$f'(x)=5x^4+1=0$, then I become unsure... any advice is greatly appreciated!
You've got it right so far- we wish to solve $5x^4+1=0$, and this will give us the location of the maximum and minimum of the function.
But consider, if we simplify $5x^4+1=0$, we get $x^4=-1/5$... This obviously has no real solutions, and so there is no point at which a maximum or minimum can exist.
However, we are given an interval over which to check- there can be no point between $-1$ and $1$ that is a maximum or minimum (b/c there are no solutions to $5x^4+1=0$), and therefore one of $1$ and $-1$ will be our maximum, and the other our minimum. So, we plug $-1$ and $1$ back into the equation: $(-1)^5 + (-1) + 1 = -1$, and $1^5 + 1 + 1 = 3$.
Hence $-1$ is our minimum, with $f(-1) = -1$, and $1$ is our maximum, with $f(1) = 3$.