What are the critical points for #f(x) = xsqrt(16 - x^2)#?

1 Answer
Jun 26, 2015

Answer:

The critical numbers are #+-2sqrt2#, and #+-4#.

Explanation:

Critical numbers for a function, #f#, are numbers in the domain #f# at which #f'(x) = 0# or #f'(x)# does not exist.

For, #f(x) = xsqrt(16 - x^2)#, the derivative is:

#f'(x) = sqrt(16-x^2) + x 1/(2sqrt(16-x^2)) *(-2x)#

# = sqrt(16-x^2)/1 - x^2/sqrt(16-x^2)#

# = (16-2x^2)/sqrt(16-x^2)#

Derivative is 0
#f'(x) = 0# when #16-2x^2 = 0#.
Which happens at #x=+-sqrt8 = +-2sqrt2#.

Both #sqrt8# and #-sqrt8# are in the domain of #f#, so they are both critical numbers.

Derivative does not exist
#f'(x)# does not exist when #16-x^2 =0#, which happens at #+-4#.
Both #4# and #-4# are in the domain of #f#, so they are both critical numbers.