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Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = ln(x^2 + 7x + 14), [−4, 1]

Sagot :

Find the stationary points of the function (where the derivative is zero):

f(x) = ln(x ² + 7x + 14)

→   f '(x) = (2x + 7) / (x ² + 7x + 14)

The denominator is always positive, since

x ² + 7x + 14 = (x + 7/2)² + 7/4 ≥ 7/4 > 0

so f '(x) = 0 when

2x + 7 = 0   →   x = -7/2

at which point f (-7/2) = ln(203/4) ≈ 3.927.

Also check the endpoints of the given domain:

f (-4) = ln(2) ≈ 0.693

f (1) = ln(22) ≈ 3.091

Then on the interval [-4, 1], we have max(f ) = ln(203/4) and min(f ) = ln(2).

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