Question

Find the inflection point of:

f(x) = e^{cosx} on [0,π], correct to 6 decimal places

I know:
f'(x) = e^(cosx)*(-sinx)
f''(x) = e^(cosx)*(-cosx) + e^(cosx)*(-sinx)(-sinx)
= e^(cosx)*(-cosx) + e^(cosx)*(sinx)^2

Then, I factor out e^(cosx) and let f''(x) = 0

e^(cosx)[(sinx)^2 - cosx] = 0

So I would solve for (sinx)^2 - cosx = 0 and find the x value for the inflection point?

Answer 1

assuming that your work is correct, and only going to the last line, solve
\[\sin^2(x)-\cos(x)=0\] by replacing \(\sin^2(x)\) by \(1-\cos^2(x)\) and solving
\[1-\cos^2(x)-\cos(x)=0\]

Answer 2

For the x value I got x = 0.904557. I printed the function and its first 2 deriv.