find the relative min and relative max of g(x) = x square root of 7 - x

Question

Also don't know how to start this one. Please help

Answer 1

g(x) = xsqr(7-x) take the derivative and set equal to zero, solve for x

g'(x) = x(1/2)/sqr(7-x) + sqr(7-x) = 0

x/2sqr(7-x) = -sqr(7-x) square both sides

x^2/4(7-x) = 7-x

x^2 = (7-x)(4)(7-x) = 4(49-14x +x^2) = 196 - 56x + 49x^2

48x^2 -56x + 196 = 0

12x^2 -14x + 49 = 0

x = 14/24 + or - (1/24)sqr(14^2 -4(12)(49)

there is no real solution, the discriminant <0

there are no local, relative extrema.

sqr(7-x) has a domain for x of 7-x>0 or x<7, for x>7 the function is undefined

this all assumes you meant f(x) =xsqr(7-x) and not (xsqr7) -x = x(sqr7 -1) which also has no local extrema, as it's a straight linear line