Question

21) Given
f(t)=∫t0x2+11x+281+cos2(x)dxf(t)=∫0tx2+11x+281+cos2(x)dx
At what value of tt does the local max of f(t)f(t) occur?

t=

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Answer 1

An extremum will occur at a value of t.

By the Fundamental Theorem of Calc

f '(t) = (t² + 11t + 28)/(1 + cos²(t)).

The numerator factors nicely making it easy to find roots. First, note that the derivative is always well defined since the denominator is always positive (it's ≥ 1).

f '(t) = (t + 4)(t + 7)/(1 + cos²(t)) ==> f '(t) = 0 if t = -4 or t = -7.

Since the denominator is always positive, it's easy to show that

f '(t) > 0 if t < -7 or t > -4, and
f '(t) < 0 if -7 < t < -4.

By the first derivative test, f takes a local maximum at t = -7