Rewrite equation:
f(x) = x sqrt(4-x^2)-x = x(4-x^2)^(1/2) - xf(x)=x√4−x2−x=x(4−x2)12−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)
