What are the local extrema, if any, of #f (x) =sqrt(4-x^2)#?

1 Answer

Answer:

The extrema of f(x) is:

  • Max of 2 at x = 0
  • Min of 0 at x = 2, -2

Explanation:

To find the extrema of any function, you carry out the following:

1) Differentiate the function
2) Set the derivative equal to 0
3) Solve for the unknown variable
4) Substitute the solutions into f(x) (NOT the derivative)

In your example of #f(x) = sqrt(4-x^2)#:

# f(x) = (4-x^2)^(1/2) #

1) Differentiate the function:

By Chain Rule**:

#f'(x) = 1/2(4-x^2)^(-1/2)*(-2x) #

Simplifying:

#f'(x) = -x(4-x^2)^(-1/2) #

2) Set the derivative equal to 0:

# 0 = -x(4-x^2)^(-1/2) #

Now, since this is a product, you can set each part equal to 0 and solve:

3) Solve for the unknown variable:

# 0 = -x # and # 0 = (4-x^2)^(-1/2) #

Now you can see that x = 0, and to solve the right side, raise both sides to the -2 to cancel out the exponent:

# 0 ^-2 = ((4-x^2)^(-1/2))^(-2)#

# 0 = 4-x^2 #
# 0 = (2-x)(2+x) #
# x = -2, 2 #

4) Substitute the solutions into f(x):

I'm not going to write out the full solution for the substitution as it is straightforward, but I'll tell you:

#f(0) = 2#
#f(-2) = 0#
#f(2) = 0#

Thus, you can see that there is an absolute maximum of 2 at x = 0, and an absolute minimum of 0 at x = -2, 2.

Hopefully everything was clear and concise! Hope I could help! :)