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f is function where f(pi/2)=17 and f ' (x)=2e^(xcosx). L(x)=line tangent to f at x=pi/2. L(2) < f(2)? JUSTIFY!

2006-10-15 08:16:29 · 1 answers · asked by :) 2 in Science & Mathematics ➔ Mathematics

1 answers

m = f"(pi/2) = 2e^0 = 2
L - 17 = 2(x - pi/2)
L = 2x + 17 - pi
L(2) = 21 - pi
In lieu of doing a numerical integration of f'(x),
L(2) > f(2) because m(2) < m(pi/2)

2006-10-15 08:40:10 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋

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