Question

Find the absolute maximum value and absolute minimum value of fx=x^2-8x+14 on the interval (5,6).

a) The absolute maximum value is

b) The absolute minimum value is

Answer 1

You have\[f(x)=x^2-8x+14,\]with\[f'(x)=2x-8.\]Setting the latter equal to zero and solving for \(x\) yields:\[0=2x-8\implies x=4.\]However, this is outside of the range \([5,6]\), and since \(f\) is a monotonically increasing function, it follows that its absolute maximum value on that interval is at \(f(6)\) and its absolute minimum value is at \(f(5)\).

Answer 2

By the way, I had to use square brackets because there is no such thing as a maximum/minimum value defined on the boundary of an open interval.

Answer 3

Not so sure about the monotonically increasing part, but it certainly is on this restriction of the domain. f(5)=-1 (minimum on the restricted domain) and f(6)=2 (the max on the restricted domain).

Answer 4

I meant on the closed interval, since \(4\) is on the left. :P

Answer 5

I figured as much.