# Determine the intervals on wich f(x)=4-x^2 is increasing and the intervals on wich it is decreasing.f(x)= 4-x^2

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In order to determine the intervals on which f(x) is an increasing or decreasing function, we'll do the first derivative test.

If the first derivative of the function is positive, then the function is increasing.

If the first derivative of the function is negative, then the function is decreasing.

We'll calculate the derivative of f(x):

f'(x) = (4-x^2)'

f'(x) = -2x

We'll put f'(x) = 0.

-2x = 0 for x = 0.

**If x<0, then -2x > 0, so f'(x) is positive, then the function is increasing over the interval (-infinite,0).**

For x>0, then -2x<0.

But f'(x) = -2x, so f'(x) is negative.

**If f'(x) is negative, the function is decreasing over the interval (0,+infinite).**

**For x = 0, the function has an extreme point, namely a maximum point.**

**f(0) = 4**

**The maximum point is: (0,4).**

f(x) = 1-x^2.

To determine the interval wher f(x) is increasing or decreasing..

We know that the if derivative is positive , then the finction is increasing.

f(x) =1-x^2.

f'(x) = -2x.

Therefore f'(x) is positive when x < 0 or when x is negative.

f'(x) = 1-x^2 is negative when x> 0 or x is positive.

So f(x) is an increaing function when x<0 or x belongs to the interval (-infinity , 0)

f(x) is decreasing as f'(x) < 0 , when x> 0. Or when x belongs to the interval (0 ,infinity).