f″(x) = sin(x28)−2cos(x)
Let's plug in each of the given critical x-values into the second derivative to see whether it is positive or negative.
f″(0) = sin(028)−2cos(0) = sin(0)−2cos(0) = 0−2(1) = −2 < 0
Since f''(0) < 0 the graph of f is concave down at x = 0 and so a local max occurs at x = 0
f″(6.949) = sin(6.94928)−2cos(6.949) ≈ −1.817 < 0 (assuming x is in radians)
Since f''(6.949) < 0 the graph of f is concave down at x = 6.949 and so a local max occurs at x = 6.949
Here's some more info about this:
https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-analytical-applications-new/ab-5-7/v/second-derivative-test
The question implies that the real part of 11−z is the same for all nonreal complex values of z such that |z|=1 . So assuming that's true, we can pick any nonreal complex z such that |z|=1 and find the real part of 11−z .
Let's pick z = 0 + 1i
Then...
11−z = 11−(0+1i) 11−z = 11−1i 11−z = 11−1i⋅1+1i1+1i 11−z = 1+1i1−i2 11−z = 1+1i1−(−1) 11−z = 1+1i2 11−z = 12+12i
The real part is 12
We can check a few more cases to see that it is 1/2 for those as well:
It seems like no matter what the angle is, the real part is 1/2