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]?

2 Answers
Jul 18, 2016

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

Explanation:

We use substitution #t=5sinx, t in [-1,5]#.

Observe that this substitution is permissible, because,

# 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#.

Jul 18, 2016

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#

Explanation:

#f(t)=tsqrt(25-t^2)#

The derivative of the function:

#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#

For the absolute maximum:

#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:

graph{xsqrt(25-x^2) [-14.4, 21.63, -5.14, 12.87]}