What are the local extrema, if any, of #f (x) =sqrt(4-x^2)#?
1 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
1) Differentiate the function:
By Chain Rule**:
Simplifying:
2) Set the derivative equal to 0:
Now, since this is a product, you can set each part equal to 0 and solve:
3) Solve for the unknown variable:
Now you can see that x = 0, and to solve the right side, raise both sides to the -2 to cancel out the exponent:
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:
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! :)