How do you find the global extreme values for #V(x) = x(10  2x)( 16  2x)# on [0,5]?
2 Answers
Answer:
Solving
Explanation:

We put the function in the polynomial form because it's easier to derivate it:
#V(x)=x(10−2x)(16−2x)=(10x2x^2)(162x)=4x^332x^220x^2+160x=4x^352x^2+160x# 
We calculate the first derivative of the function:
#V'(x)=12x^2104x+160=4(3x^226x+40)# 
Now we can find the global extreme values solving
#V'(x)=0# :
#3x^226x+40=0#
#x_1=2# and#x_2=20/3# 
We can also determine the nature of these two points by calculating the sign of the derivative:
#V(x)<0# if#x<20/3 and x>2#
#V(x)>=0# if#x>=20/3 or x<=2#
So the function#V(x)# grows until#x=2# then it decease until#x=20/3# and then it grows again. So#x_1# is a relative maximum and#x_2# is a relative minimum. 
We can verify it by seeing the graph of this function:
graph{4x^352x^2+160x [3, 14, 200, 200]}
Answer:
This is a closed interval type problem. The global extrema occur at critical numbers in the interval or at endpoints of the interval.
Explanation:
The function can be further factored, to get:
# = x[(2)(x5)]9(2)(x8)]#
# = 4x(x5)(x8)#
We need the critical numbers, so we need the derivative of
So,
The critical numbers (in
But
.
The extrema occur at either the endpoints, (
Answer:
The maximum value is
The minimum is