Suppose
f(x)=∫ (from 0 to x) (t^2-36)/2+cos^2(t) dt.
For what value(s) of x
does f(x) have a local maximum?
well, df/dx = (x^2-36)/2 + cos^2(x)
where is that zero?
which is a max?
How do we find the zero is it with the denominator or the numerator ?
well, there's no algebraic way to do it that I know of. Best to check the graph and approximate it. You know that x^2-36 is zero at ±6. cos^2(x) is positive there, so x will be on the inside of the interval (-6,6). So you can check nearby values and approximate it as closely as you want.
Now you have to decide which is a max and which is a min when f'=0. You know that at a maximum, the slope changes from positive to negative (top of a hill). So, since f'(-6) > 0 and f'(-5) < 0, the value near -6 is a maximum. The graph of f' is at
http://www.wolframalpha.com/input/?i=(x%5E2-36)%2F2+%2B+cos%5E2(x)
To find the value(s) of x for which the function f(x) has a local maximum, we need to find the critical points of the function and then determine whether each critical point is a maximum or minimum.
1. Take the derivative of the function f(x) with respect to x. This will help us find the critical points.
f'(x) = (x^2 - 36)/2 + cos^2(x)
2. Set f'(x) equal to zero and solve for x to find the critical points.
(x^2 - 36)/2 + cos^2(x) = 0
3. To solve this equation, we first simplify it.
x^2 - 36 + 2cos^2(x) = 0
4. Rearrange the equation to isolate cos^2(x) term.
2cos^2(x) = 36 - x^2
5. Divide both sides by 2.
cos^2(x) = (36 - x^2)/2
6. Take the square root of both sides.
cos(x) = ±sqrt((36 - x^2)/2)
7. Now, we have two separate equations for positive and negative square roots.
cos(x) = sqrt((36 - x^2)/2)
cos(x) = -sqrt((36 - x^2)/2)
8. To find the critical points, solve each equation separately.
For cos(x) = sqrt((36 - x^2)/2):
Take the inverse cosine of both sides.
x = arccos(sqrt((36 - x^2)/2))
For cos(x) = -sqrt((36 - x^2)/2):
Take the inverse cosine of both sides.
x = arccos(-sqrt((36 - x^2)/2))
9. Solve each equation for x to find the critical points. You can use numerical methods or graphing calculator/software to approximate the solutions.
10. Once you have the critical points, you can use the second derivative test to determine whether each critical point is a local maximum or minimum.
a. Take the second derivative of f(x):
f''(x) = d^2/dx^2 [(x^2 - 36)/2 + cos^2(x)]
f''(x) = -xsin(x) + 2cos(x)sin(x)
b. Substitute each critical point obtained in step 9 into f''(x).
c. If f''(x) is positive at a critical point, then it is a local minimum.
If f''(x) is negative at a critical point, then it is a local maximum.
If f''(x) is zero at a critical point, then the test is inconclusive.
By following these steps, you can find the value(s) of x for which f(x) has a local maximum.