SOLUTION: Consider f(x) = x sqrt(x + 2) and g(x) = x/ sqrt(5 − x) .(a) Find the domains of the functions f and g. (b) Find the derivatives of the functions f and g in part (a). (c)

Question 1032808: Consider
f(x) = x sqrt(x + 2) and g(x) = x/ sqrt(5 − x)
.(a) Find the domains of the functions f and g.
(b) Find the derivatives of the functions f and g in part (a).
(c) Find the values of x so that (i) f'(x) = 0, (ii) g'(x) = 0

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Domain of x(sqrt(x+2)) are all x greater than or equal to -2. Any x less than that gives sqrt (number <0)
domain of x/sqrt(5-x) is x less than 5.

f(x)=x*(x+2)^(1/2)
f'(x)=x*(1/2)(x+2)^(-1/2)+(x+2)^(1/2)
set that equal to 0.
sqrt(x+2)=-x(1/2sqrt(x+2))
square both sides
x+2=x^2/4(x+2)
4(x+2)(x+2)=x^2
4x^2+16x+16=x^2
3x^2+16x+16=0
(3x+4)(x+4)=0
x=-4/3.-4, but -4 doesn't work in original derivative.
x=-4/3 for derivative = 0. Notice that the tangent to the function at -4/3 has 0 slope.

==============================
g(x)=x*(5-x)^(1/2)
g'(x)=x*(1/2)(5-x)^(-1/2)+(5-x)^(1/2)
set it equal to 0
0=x(1/2)(5-x)^(-1/2)+(5-x)^(1/2)
-(5-x)^(1/2)=x*(1/2)(1/sqrt(5-x))
square both sides
5-x=x^2(1/4(5-x))
4(5-x)^2=x^2
4x^2-40x+100=x^2
3x^2-40x+100=0
(3x-10)(x-10)=0
=10/3,10
But x cannot equal these values, so first derivative will not ever equal 0. In the graph, there is no place where the slope of the tangent line is 0.

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SOLUTION: Consider f(x) = x sqrt(x + 2) and g(x) = x/ sqrt(5 &#8722; x) .(a) Find the domains of the functions f and g. (b) Find the derivatives of the functions f and g in part (a). (c)

SOLUTION: Consider f(x) = x sqrt(x + 2) and g(x) = x/ sqrt(5 − x) .(a) Find the domains of the functions f and g. (b) Find the derivatives of the functions f and g in part (a). (c)