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
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
2 answers
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
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
1 answer