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.
To find and classify the relative maxima and minima of the function f(x) = ∫[0,x] (t^2-4/(1+cos(t)^2)) dt, you need to follow these steps.
Step 1: Find the critical points
To find the critical points, you need to find the values of x where the derivative of f(x) is equal to zero or undefined. Start by finding the derivative of f(x):
f'(x) = (d/dx)∫[0,x] (t^2-4/(1+cos(t)^2)) dt
To differentiate the integral, you can use the Fundamental Theorem of Calculus, which states that if F(x) is an antiderivative of f(x), then the derivative of the definite integral is F(b) - F(a), where [a,b] is the interval of integration. Applying this theorem, you can find the derivative:
f'(x) = (x^2-4/(1+cos(x)^2)) - (0^2-4/(1+cos(0)^2))
Remember that the derivative of a constant is zero.
Simplifying further:
f'(x) = x^2-4/(1+cos(x)^2) - 4
Now, set this derivative equal to zero and solve for x:
x^2-4/(1+cos(x)^2) - 4 = 0
Step 2: Solve for x
To solve the equation x^2-4/(1+cos(x)^2) - 4 = 0, you'll need to use numerical methods since it does not have a simple algebraic solution. One common numerical method is to use a graphing calculator or an online tool like Wolfram Alpha to find approximate solutions.
When you typed the keyword 'local maximum x^2-4/(1+cos(x)^2)' into Wolfram, it gave you a list of local maxima. Wolfram likely used numerical methods to approximate the values of x where the derivative is equal to zero. The values it provided (such as +/-4.99 and +/-7.99) are likely close approximations to the actual values.
Step 3: Classify the critical points
To classify each critical point, you'll need to examine the behavior of the function around each point. You can do this by looking at the values of the second derivative f''(x) at each critical point.
f''(x) = (d^2/dx^2)∫[0,x] (t^2-4/(1+cos(t)^2)) dt
Again, you can use the Fundamental Theorem of Calculus to differentiate the integral:
f''(x) = (x^2-4/(1+cos(x)^2)) - (0^2-4/(1+cos(0)^2))
Simplifying:
f''(x) = x^2-4/(1+cos(x)^2)
Evaluate f''(x) at each critical point you found in Step 2 to determine if it's a relative maximum or minimum. If f''(x) > 0, it is a relative minimum. If f''(x) < 0, it is a relative maximum.
Note that if the second derivative f''(x) is undefined at a critical point, you'll need to use additional methods (e.g., the first derivative test or the sign chart method) to determine the nature of the critical point.
Overall, finding and classifying relative maxima and minima requires finding critical points and analyzing the behavior using the second derivative, and in cases without an algebraic solution, numerical methods can be used to approximate the values.