How do you find local maximum value of f using the first and second derivative tests: #f(x)=x+sqrt(2-x)#?

Question

How do you find local maximum value of f using the first and second derivative tests:  #f(x)=x+sqrt(2-x)#?

Isabella Ackerman · Accepted Answer

See below.

Domain of #f# is #(-oo,2]# #f'(x) = 1-1/(2sqrt(2-x)) = (2sqrt(2-x)-1)/(2sqrt(2-x))# #f'# is undefined at #2# and is #0# at #7/4#. So the critical numbers are #2# and #7/4#. (Some do not apply the term "critical number" at an endpoint of the domain.) Using the First Derivative Test On #(-oo,7/4)# we have #f'(x) > 0# and on #(7/4,2)#, #f'(x) < 0#, so #f(7/4) = 9/4# is a local maximum and #f(2) = 2# is a local minimum. (Some do not apply the term "local minimum" at an endpoint of the domain. They would disagree with my answer about that.) Using the Second Derivative Test (for local extrema) #f''(x) = -1/(4sqrt(2-x)^3)# #f''(7/4) < 0#, so #f(7/4) = 9/4# is a local maximum. #f''(2)# does not exist, so we cannot use the second derivative test at #x=2#

Elijah Abram · Answer

undefined To find the local maximum value of $ f(x) = x + \sqrt{2 - x} $ using the first and second derivative tests:

1. Find the first derivative of $ f(x) $ and set it equal to zero to find critical points.
2. Find the second derivative of $ f(x) $ to determine the concavity of the function at the critical points.
3. Use the first derivative test to determine whether the critical points are local maxima, minima, or neither.
4. Use the second derivative test to confirm the nature of the critical points found in step 3.

First derivative of $ f(x) $:
$$ f'(x) = 1 - \frac{1}{2\sqrt{2 - x}} $$

Set $ f'(x) = 0 $ to find critical points:
$$ 1 - \frac{1}{2\sqrt{2 - x}} = 0 $$
$$ \frac{1}{2\sqrt{2 - x}} = 1 $$
$$ \sqrt{2 - x} = 2 $$
$$ 2 - x = 4 $$
$$ x = -2 $$

Second derivative of $ f(x) $:
$$ f''(x) = \frac{1}{4(2 - x)^{3/2}} $$

Substitute $ x = -2 $ into $ f''(x) $ to determine the concavity:
$$ f''(-2) = \frac{1}{4(2 - (-2))^{3/2}} = \frac{1}{4(4)^{3/2}} > 0 $$

Since $ f''(-2) > 0 $, the function is concave up at $ x = -2 $, indicating a local minimum.

Therefore, the local maximum value of $ f(x) $ does not occur at $ x = -2 $. There is no local maximum value for the given function.