
Maxima and Minima 7-11
Finding the critical values:
To find the critical values, we set y′ = 0.
Thus, y
x x
x
' =
− +
−
=
2
2
20 64
10
0
( )
.
Thus, x
2
- 20x + 64 = 0 Þ (x - 4) (x - 16) = 0.
That is, x = 4 or x = 16 are the critical points.
Testing the first derivative for maximum and minimum values:
Consider x = 4.
y
x x
x
' =
− +
−
2
2
20 64
10( )
.
We see that for x slightly less than 4, say 3, y′ is positive. Similarly, for x slightly greater than 4, say
5, y′ is negative.
Thus, as x changes from values slightly less than 4 to values slightly more than 4, y′ changes from
positive to negative.
Hence, the original function has a maximum value at x = 4 and the maximum ...