Consider the function f(x) = e ^-x on the value [0,ln2] Find the "c" value guaranteed by the mean value theorem, or if the mean value theorem conditions are not met state why

Answer 1

f(ln2) = 1/2

f(0) = 1

The slope of the secant line joining the endpoints of the interval is (1-.5)/(-ln2)...

Somewhere between 0 and ln 2 there exists a "c" such that the slope of the tangent line at (c, f(c) is equal to the slope of the secant line (since f is continuous on the closed interval [0, ln2] and differentiable on (0,ln2) -- MVT.

f'(x) = -1/e^x

Solve the equation -1 / e^x = .5/(-ln2)

-1/e^x = -1/(2ln2) = -1/ln4

Therefore e^x = ln4

and, x = ln(ln4)

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desmos.com/calculator/p2nnttjlik