Increasing and Decreasing Functions

g(x)= x(sqrt(10-x^2))

Where is g increasing?

Where is g decreasing?



thanks

g is decreasing when the derivitve of the function is negative and g is increasing when the derivitve of the function is positive. This makes sense, because the derivitive represents the slope, so when the slope is negative the function is clearly decreasing, and when the slope is positive the function is increasing.

g(x) = x * (10-x^2)^.5
g'(x) = x * .5(-2x) (10-x^2)^-.5 + (10-x^2)^.5

now, we must find out when g is positive or negative. How do we do this? well, find when g is zero, and test any x value in between any values of x.

in this case,

0 = -x^2 (10-x^2)^-.5 + (10-x^2)^.5
combine the fractions (remember that raising something to a negative power is the same as divided by the positive of that power)

0 = (-x^2 + (10-x^2)^.5 * (10-x^2)^.5)/(10-x^2)^.5

the function must equal zero when the numerator equals zero.

-x^2 + 10-x^2 = 0
10 = 2x^2

x = +- (5)^.5
that means we must check a point greater than 5^.5 and less than -5^.5 and in between the two.
I will leave this to you. make sure to choose points in the domain of g (no negative square roots, which means that you can not choose an x that is greater than (10)^.5.

you should get that it decreases before -(5)^.5 , increases in between, and decreases after 5^.5

Hope this helps!