Find and classify the relative maxima and minima of f(x) if f(x)= defint a=0 b=x
function= t^2-4/(1+cos(t)^2) dt
x^2-4/(1+cos(x)^2)= 0
x^2-4=0
x^2=4
x= +/- 2
So I got relative maximum as -2 and 2. And relative minimum as zero. However, when I graph it on Wolfram, it gives me more maxima like +/-4.99, +/-7.999, etc. How did they get those values? Can someone please explain that to me? Thank you for your time.
Sorry I wrote the wrong variable in the last posting (this is a correction).
I got +/-4.99 and +/-7.99 when I typed the keyword 'local maximum x^2-4/(1+cos(x)^2)'
into the equation box. It just gave me a list of maxima.
You should write the function (derivative of the integral) as:
(x^2-4)/(1+cos(x)^2)
this is zero for x = +/- 2.
you don't want the minima and maxima of (x^2-4)/(1+cos(x)^2)
That is not f(x). It is f'(x). So, the minima and maxima of f(x) occur when (x^2-4)/(1+cos(x)^2) = 0.
To find the relative maxima and minima of the function f(x) = ∫[from 0 to x] t^2-4/(1+cos(t)^2) dt, you correctly started by solving the equation f'(x) = 0.
However, the mistake you made was in solving x^2-4/(1+cos(x)^2) = 0. The expression x^2-4 is not equal to zero, so that step is incorrect.
To find the critical points and classify the relative extrema of f(x), you need to find where f'(x) = 0 or where it does not exist. Let's go through it step by step:
Step 1: Calculate the derivative of f(x) using the Fundamental Theorem of Calculus:
f'(x) = (d/dx) ∫[from 0 to x] t^2-4/(1+cos(t)^2) dt
Step 2: Apply the Fundamental Theorem of Calculus to differentiate the integral:
f'(x) = x^2-4/(1+cos(x)^2)
Step 3: Set f'(x) equal to zero and solve for x:
x^2-4/(1+cos(x)^2) = 0
Step 4: Simplify the equation:
Multiply both sides by (1+cos(x)^2):
x^2 - 4 = 0
Step 5: Solve for x:
x^2 = 4
x = ±2
So far, you have correctly found the critical points x = -2 and x = 2. These are the only critical points for this function.
To classify the nature of these critical points as relative maxima or minima, you need to use the second derivative test. Let's proceed with that:
Step 6: Calculate the second derivative of f(x) by differentiating f'(x) with respect to x:
f''(x) = (d/dx) [x^2-4/(1+cos(x)^2)]
Step 7: Simplify the expression:
f''(x) = 2x/(1+cos(x)^2) - 2x*sin(2x)/(1+cos(x)^2)^2
Step 8: Evaluate the second derivative at x = -2 and x = 2:
For x = -2:
f''(-2) = 2(-2)/(1+cos(-2)^2) - 2(-2)*sin(2(-2))/(1+cos(-2)^2)^2
For x = 2:
f''(2) = 2(2)/(1+cos(2)^2) - 2(2)*sin(2(2))/(1+cos(2)^2)^2
To classify the critical points as relative maxima, relative minima, or neither, you need to analyze the sign of the second derivative at each critical point.
If f''(x) > 0 at a critical point x, then it is a relative minimum.
If f''(x) < 0 at a critical point x, then it is a relative maximum.
If f''(x) = 0 at a critical point x, then the test is inconclusive.
Unfortunately, the expression for the second derivative, f''(x), is complicated. To determine the nature of the critical points more accurately, you can graph the function or use numerical methods to find additional critical points.
It seems like when you searched for "local maximum x^2-4/(1+cos(x)^2)" on Wolfram, it might have used numerical methods to find other maxima.
To find the relative maxima and minima of the function f(x) = ∫[a=0, b=x] (t^2 - 4/(1 + cos(t)^2)) dt, we need to first find the critical points by setting the derivative of f(x) equal to zero.
Now, to differentiate f(x) with respect to x, we can use the Fundamental Theorem of Calculus. The derivative of f(x) is given by:
f'(x) = (d/dx) (∫[a=0, b=x] (t^2 - 4/(1 + cos(t)^2)) dt).
Using the chain rule, we get:
f'(x) = x^2 - 4/(1 + cos(x)^2) * (d/dx) (x) = x^2 - 4/(1 + cos(x)^2).
Setting f'(x) equal to zero:
x^2 - 4/(1 + cos(x)^2) = 0.
Multiplying through by (1 + cos(x)^2):
x^2(1 + cos(x)^2) - 4 = 0.
Expanding:
x^2 + x^2cos(x)^2 - 4 = 0.
Rearranging:
x^2cos(x)^2 = 4 - x^2.
Dividing through by x^2:
cos(x)^2 = (4 - x^2)/x^2.
We know that cos(x)^2 lies between 0 and 1, so:
0 ≤ (4 - x^2)/x^2 ≤ 1.
Now, we can classify the relative maxima and minima by evaluating the function at the critical points, as well as the endpoints of the interval [a=0, b=x].
1. Critical Points:
To find the critical points, we need to solve the equation cos(x)^2 = (4 - x^2)/x^2.
Simplifying further, we get:
cos(x)^2 = 4/x^2 - 1.
Multiplying through by x^2, we have:
x^2 * cos(x)^2 = 4 - x^2.
We already know that x = ±2 are solutions to this equation.
Now, we can use a numerical method (such as the Newton-Raphson method or a graphing utility) to find the approximate values of the other critical points that you mentioned, such as ±4.99, ±7.99, and more. These values might not have simple closed-form solutions and require numerical approximation techniques.
2. Endpoints:
Evaluate the function at the endpoints of the interval [a=0, b=x]. Since the function is defined as an integral from a=0 to b=x, as x approaches infinity, the integral value will also approach infinity. Similarly, as x approaches negative infinity, the integral value will approach negative infinity.
Thus, there are no relative maxima or minima at the endpoints of the interval.
Overall, you correctly found the relative maxima as x = ±2. However, the additional values you mentioned (±4.99, ±7.99, etc.) could be obtained through numerical methods or graphing utilities, which approximate the other critical points.