mean value theorem backwards ? :)
df/dx = 1/(x-3)
f(9) = ln(6) = 1.79
f(5) = ln(2) = .69
f(9)-f(5) = 1.0986
divide by 4
= .27465
1/(x-3) = .27465
ln(x-3) =
x-3 = 1/.27465
x = 3 + 3.64
x = 6.64
check
1/(6.64-3) = .2747
for f(x)= ln(x-3) on the interval [5,9], find c so that f'(c)= (f(b)-f(a))/(b-a)
1 answer