How do you find the absolute maximum and absolute minimum values of f on the given interval: #f(t) =t sqrt(25t^2)# on [1, 5]?
2 Answers
Reqd. extreme values are
Explanation:
We use substitution
Observe that this substitution is permissible, because,
which holds good, as range of
Now,
Since,
Therefore, reqd. extremities are
Find the monotony of the function from the derivative's sign and decide which local maximum/minimums are the biggest, smallest.
Absolute maximum is:
Absolute minimum is:
Explanation:
The derivative of the function:

The numerator has two solutions:
#t_1=sqrt(12.5)=3.536#
#t_2=sqrt(12.5)=3.536#
Therefore, the numerator is:
Negative for#t in(oo,3.536)uu(3.536,+oo)#
Positive for#t in(3.536,3.536)# 
The denominator is always positive in
#RR# , since it's a square root.
Finally, the range given is#[1,5]#
Therefore, the derivative of the function is:
 Negative for
 Positive for
This means the graph firstly goes up from
Absolute maximum is
For the absolute maximum:
Therefore,
You can see from the graph below that this is true. Just ignore the area left of
graph{xsqrt(25x^2) [14.4, 21.63, 5.14, 12.87]}