Consider the function below.
A(x) = x squt (x + 12)
Find the local maximum value
I tried to plug in f(-12) in to A(x) results = 0
but it aint correct
for a local maximum value you need
the derivative of x √(x + 12) to be zero
A(x) = x(x + 12)^(1/2)
A'(x) = x (1/2)(x+12)^(-1/2) + (x+12)^(1/2)
= (1/2)(x+12)^(-1/2) [ x + 2(x+12)) = 0
well, (1/2)(x+12)^(-1/2) can never be zero, so
x + 2(x+12) = 0
3x = -24
x = -8
then A(-8) = -8√(-8+12) = -16
Now look at the graph:
https://www.wolframalpha.com/input/?i=graph+y+%3D+x+%E2%88%9A%28x+%2B+12%29
You can see that (-8,-16) is a minimum, not a maximum.
We could show this by taking the 2nd derivative and showing that for x=-8 the original function
would be concave up.
You should have included the domain of x ≥ -12
of course A(-12) will not be the maximum, since it is negative, and A(x) > 0 for x>0
So, since you're doing calculus, recall that A(x) will have a maximum or minimum where A'(x) = 0. Using the product rule,
A = x√(x+12)
A' = √(x+12) + x/(2/(x+12)) = (2(x+12)+x)/(2√(x+12)) = 3(x+8)/(2√(x+12))
Now, A' = 0 when x = -8
Clearly this cannot be a maximum, so it must be a minimum.
So A(x) has no maximum unless you restrict the domain.
See the graph at
https://www.wolframalpha.com/input/?i=x%E2%88%9A%28x%2B12%29
To find the local maximum value of the function A(x) = x√(x + 12), you would need to differentiate the function and find the critical points where the derivative is equal to zero or undefined. The critical points can help us determine if there are local maximum or minimum values.
Let's find the derivative of A(x) using the power rule and the chain rule:
A'(x) = d/dx (x√(x + 12))
= (√(x + 12)) + x * (1/2)(x + 12)^(-1/2)
To find the critical points, we need to solve the equation A'(x) = 0:
(√(x + 12)) + x * (1/2)(x + 12)^(-1/2) = 0
Since we have a square root term, let's multiply both sides of the equation by (√(x + 12)) to get rid of the square root:
(√(x + 12))^2 + x * (1/2) = 0
(x + 12) + (x/2) = 0
2(x + 12) + x = 0
2x + 24 + x = 0
3x + 24 = 0
3x = -24
x = -8
Now we have a critical point at x = -8. To determine if it is a local maximum or minimum, we can consider the second derivative.
A''(x) = d^2/dx^2 (x√(x + 12))
= d/dx (√(x + 12)) + (1/2)(x + 12)^(-1/2)
= (1/2)(x + 12)^(-1/2)
Now we substitute the critical point x = -8 into the second derivative:
A''(-8) = (1/2)(-8 + 12)^(-1/2)
= (1/2)(4)^(-1/2)
= (1/2)(1/2)
= 1/4
Since the second derivative is positive at x = -8, the critical point represents a local minimum. Therefore, there is no local maximum value for the given function A(x) = x√(x + 12).