f(t) = \int_0^t \frac{x^2+11 x+24}{ 1+\cos^2(x)} dx
At what value of t does the local max of f(t) occur?
t = ?
f'(t) = (t^2+11t+24)/(1+cos^2(t))
f'=0 when t^1+11t+24 = 0
That is, when t = -8 or -3
f"(t) = 2t+11
f"(-8) < 0, so the max is there.
To find the local maximum of the function f(t), we need to find the critical points of the function and determine which one gives the maximum value.
Step 1: Find the first derivative of f(t).
To find the derivative of f(t), we need to apply the Fundamental Theorem of Calculus and the Chain Rule. The derivative of f(t) with respect to t is given by:
f'(t) = (d/dt) [∫₀ₜ (x² + 11x + 24)/(1 + cos²(x)) dx]
Step 2: Simplify the expression of f'(t).
To find the derivative of the given integral, we can employ the Leibniz Rule or the Newton-Leibniz Rule. The integral is a function of t, so we can differentiate it with respect to t. Here is the simplified version of f'(t):
f'(t) = (t² + 11t + 24)/(1 + cos²(t))
Step 3: Set f'(t) equal to zero and solve for t.
To find the critical points, we set the first derivative f'(t) equal to zero and solve the resulting equation:
(t² + 11t + 24)/(1 + cos²(t)) = 0
However, since the denominator (1 + cos²(t)) is always positive, the numerator (t² + 11t + 24) must be zero to satisfy the equation.
Using factoring or the quadratic formula, we can solve the equation t² + 11t + 24 = 0 to find the values of t that make the first derivative equal to zero. The solutions will give us the potential critical points.
Step 4: Determine the local maximum.
After obtaining the values of t that make the first derivative equal to zero, we evaluate f(t) at those critical points and select the t-value that gives the maximum value of f(t). The t-value corresponding to the local maximum of f(t) will be the answer.
By following these steps, you can find the value of t at which the local maximum of f(t) occurs.