What are the absolute extrema of # f(x)= xsqrt(4-x^2) -x# in #[-1,2]#?

1 Answer
Feb 24, 2017

Answer:

Relative max. at #x = 1/2 sqrt(1/2(15-sqrt(33)) #
Relative min. at right endpoint: #(2, -2)#

Explanation:

Rewrite equation:
#f(x) = x sqrt(4-x^2)-x = x(4-x^2)^(1/2) - x#

Find the first derivative using the product rule #(uv)' = u*v' + v*u'# and Power rule: #(u^n)' = n u^(n-1)u':#

Let #u = x#, #u' = 1#
and #v = (4-x^2)^(1/2)#, #v' = 1/2(4-x^2)^(-1/2)(-2x) = (-x)/sqrt(4-x^2)#

#f(x)' = x * (-x)/sqrt(4-x^2) + sqrt(4-x^2) *(1) - 1#

Find a common denominator and simplify:

#f(x)' = (-x^2)/sqrt(4-x^2) + (sqrt(4-x^2)sqrt(4-x^2))/sqrt(4-x^2) - sqrt(4-x^2)/sqrt(4-x^2) = (-x^2+4-x^2 - sqrt(4-x^2))/sqrt(4-x^2) = (4-2x^2-sqrt(4-x^2))/sqrt(4-x^2)#

Find the critical numbers by setting #f(x)' = 0# and simplifying:
#4-2x^2-sqrt(4-x^2) = 0#;
#4-2x^2 = sqrt(4-x^2)#
#(4-2x^2)^2 = 4-x^2#
#16-16x^2+4x^4 = 4-x^2#
#4x^4-15x^2+12 = 0#

Let #z = x^2#: #4z^2-15z+12 = 0#
Use the quadratic equation to find values of #a#:

#a = (15+-sqrt(33))/8#

Since #x = +-sqrt(a), x = +-sqrt((15+-sqrt(33))/8) = +-sqrt((15+-sqrt(33))/(2sqrt(4)))#

So #x = +-1/2 sqrt(1/2(15+-sqrt(33))#; #x~~+-1.0756, +-1.6103#

Use the first derivative test to see which critical numbers are relative maximums and relative minimums.

Intervals based on endpoints:
#[-1, 1.0756), (1.0756, 1.61030), (1.61030, 2]#
#f'(0) > 0; f'(1.5) < 0; f'(1.8) < 0#

Relative max. at #x = 1/2 sqrt(1/2(15-sqrt(33)) #
Relative min. at right endpoint: #(2, -2)#