What are the critical points of # f(x) = x*sqrt(8-x^2) #?

1 Answer
Feb 28, 2018

Answer:

x= #+-2, +-2sqrt(2)#

Explanation:

To find the critical points of a function, find where the first derivative equals 0.
For #f(x)# stated above, #f'(x) = (8-2x^2)/sqrt(8-x^2)#

You need to use the product rule for derivatives: #u * dv + v * du#
where in this case, #u = x#, and #v = sqrt(8-x^2)#

#du# is the derivative of #u# and the derivative of #x= 1# so #du = 1#.

#dv# is the derivative of #v#; the derivative of #sqrt(8-x^2)# requires the use of the Chain Rule because you are taking the derivative of the function of a function (the square root of #(8-x^2)#).
It is easier to represent #sqrt((8-x^2))# using exponents:
#sqrt((8-x^2)) = (8-x^2)^(1/2)#.

Now use the Chain Rule to find the derivative of #v#:
#d/dx((8-x^2)^(1/2)) = 1/2(8-x^2)^(-1/2)*-2x#
(Don't forget to find the derivative of the inside function --
that's where the #-2x# comes from!)

Clean this up so #dv= -x/sqrt(8-x^2)#

OKAY, put it all together now:

#u*dv + v*du =#
#x*-x/sqrt(8-x^2) + sqrt(8-x^2) * 1#
# = -x^2/sqrt(8-x^2) + sqrt(8-x^2)#

Combine these two fractions into one by getting common denominators and you get #f'(x) = (8-2x^2)/sqrt(8-x^2)#

Critical values occur when #f'# = 0 or undefined (i.e. the
numerator = 0 and/or denominator = 0).
Setting the numerator = 0 and solving results in #x=+-2#
Setting the denominator = 0 and solving results in #x=+-2sqrt(2)#

If you use Symbolab, you can quickly get these answers, I hope my explanation helped! :)