Consider the function:

Question

f(t)= integral from 0 to t of (x^2+11x+28)/(1+cos^2(x))

At what value of t does the local max of f(t) occur?

t=

Answer 1

The derivative of f(t) = (t²+11t+28)/(1+cos²(t)).

That equals 0 when the numerator is zero. Note that the denominator is always positive.

The numerator equals 0 when t = -7 or t = -4. Using a "sign" diagram, determine that f'(t) is negative to the left of -7 and positive just to the right of -7, so there is a relative max at t = -7.

Turns out there is a relative min at t = -4.

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Answer 2

The only critical points of f are at x=-4 and x=-7.

Use the messy 2nd derivative test to decide if either is a maximum.