How do you find the absolute maximum and absolute minimum values of f on the given interval: #f(t) =t sqrt(25-t^2)# on [-1, 5]?

Question

William Abdalla · Accepted Answer

Reqd. extreme values are #-25/2 and 25/2#.

We use substitution #t=5sinx, t in [-1,5]#. Note that this substitution is allowed due to # t in [-1,5] rArr -1<=t<=5rArr -1<=5sinx<=5# #rArr -1/5<=sinx<=1#, which holds good, as range of #sin# fun. is #[-1,1]#. Now, #f(t)=tsqrt(25-t^2)=5sinx*sqrt(25-25sin^2x)# #=5sinx*5cosx=25sinxcosx=25/2(2sinxcosx)=25/2sin2x# Since, #-1<=sin2x<=1 rArr -25/2<=25/2sin2x<=25/2# #rArr -25/2<=f(t)<=25/2# Therefore, reqd. extremities are #-25/2 and 25/2#.

Christopher Allers · Answer

Find the monotony of the function from the derivative's sign and decide which local maximum/minimums are the biggest, smallest.

Absolute maximum is:
#f(3.536)=12.5#

Absolute minimum is:
#f(-1)=-4.899# #f(t)=tsqrt(25-t^2)# The function's derivative: #f'(t)=sqrt(25-t^2)+t*1/(2sqrt(25-t^2))(25-t^2)'# #f'(t)=sqrt(25-t^2)+t*1/(2sqrt(25-t^2))(-2t)# #f'(t)=sqrt(25-t^2)-t^2/sqrt(25-t^2)# #f'(t)=sqrt(25-t^2)^2/sqrt(25-t^2)-t^2/sqrt(25-t^2)# #f'(t)=(25-t^2-t^2)/sqrt(25-t^2)# #f'(t)=(25-2t^2)/sqrt(25-t^2)# #f'(t)=2(12.5-t^2)/sqrt(25-t^2)# #f'(t)=2(sqrt(12.5)^2-t^2)/sqrt(25-t^2)# #f'(t)=2((sqrt(12.5)-t)(sqrt(12.5)+t))/sqrt(25-t^2)# 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 #t in[-1,3.536)# - Positive for #t in(3.536,5)# This means the graph firstly goes up from #f(-1)# to #f(3.536)# and then goes down to #f(5)#. This makes #f(3.536)# the absolute maximum and the biggest value of #f(-1)# and #f(5)# is the absolute minimum. Absolute maximum is #f(3.536)#: #f(3.536)=3.536sqrt(25-3.536^2)=12.5# Regarding the highest possible value: #f(-1)=-1sqrt(25-(-1)^2)=-4.899# #f(5)=5sqrt(25-5^2)=0# Therefore, #f(-1)=-4.899# is the absolute minimum. You can see from the graph below that this is true. Just ignore the area left of #-1# since it's out of the domain: plot{xsqrt(25-x^2) [-14.4, 21.63, -5.14, 12.87]}

Charles Anctil · Answer

undefined To find the absolute maximum and absolute minimum values of $ f(t) = t \sqrt{25 - t^2} $ on the interval $[-1, 5]$, you follow these steps:

1. Find the critical points of $ f(t) $ within the interval by setting its derivative equal to zero and solving for $ t $.
2. Evaluate $ f(t) $ at the critical points obtained in step 1, and also at the endpoints of the interval.
3. Compare the values obtained in step 2 to determine the absolute maximum and absolute minimum values of $ f(t) $ on the interval.

The critical points of $ f(t) $ are found by solving the equation $ f'(t) = 0 $, where $ f'(t) $ is the derivative of $ f(t) $ with respect to $ t $. The derivative of $ f(t) $ is calculated using the product rule.

After finding the critical points and evaluating $ f(t) $ at those points and the endpoints of the interval, you compare the values to determine which one is the absolute maximum and which one is the absolute minimum.