Show that the derivative of a function at a point is greater than zero.

Question

Given $f(x) = 4x^3-6x^2\cos(2a) + 3x\sin(2a)\sin(6a) + \sqrt{\ln(2a-a^2)}$ then show that $f'\Bigl(\frac{1}{2}\Bigr)>0.$

The obvious thing to do here is to differentiate the function,
(Since nothing is mentioned about 'a', I'll assume it to be constant. At least that's what my instructor said)

$f'(x) = 12x^2-12x\cos(2a) + 3\sin(2a)\sin(6a)$

$f'\Bigl(\frac{1}{2}\Bigr) = 3-6\cos(2a) + 3\sin(2a)\sin(6a).$

How do you prove, that this is greater than zero? I tried using the range of $\cos(x)$ and $\sin(x)$, but got stuck.

Another thing I'd like to mention that this question was given to me as a multiple choice question. The options were,

A) $f(x)$ is not defined at $x = 1/2$
B) $f'(x)$ is not defined at $x = 1/2$
C) $f\Bigl(\frac{1}{2}\Bigr)<0$
D) $f\Bigl(\frac{1}{2}\Bigr)>0$

Since we don't know what 'a' is, the answer according to me should be B). But the answer given to me was D), which my instructor said was correct, and that 'a' should be treated as a constant.

Any help would be appreciated.

Answer 1

If you execute the Mathematica command Plot[3-6Cos[2a]+3Sin[2a]Sin[6a],{a,0,2 Pi}], you will find that the result is sometimes positive, sometimes negative, depending on $a$. The derivative is defined, but it's indeterminant because you don't know what $a$ is. My answer would be "None of the above."

[EDIT]: In order for the logarithm and square root to be defined and non-complex, you do know something about $a:$ you must have $2a-a^2\ge 1,$ the solution of which is $a=1$. For this value of $a$, $f'(1/2)>0$.

Answer 2

This is an extremely poorly worded question, assuming you have not left out any part of it. The correct answer is actually either both (A) and (B) or (D), depending on the value of $a$.

First of all, $f(x)$ may not even be defined for any values of $x$, depending on what $a$ is. In order for the last term $\sqrt{\ln(2a-a^2)}$ to be defined, $2a-a^2$ must be at least $1$ so that $\ln(2a-a^2)$ is defined and nonnegative. Since $2a-a^2-1=-(a-1)^2$, the only way that can happen is if $a=1$. So if $a\neq 1$, then $f(x)$ is never defined, and in particular $f(1/2)$ and $f'(1/2)$ are not defined, so both (A) and (B) are correct answers.

On the other hand, if $a=1$, then you can simply plug that in and evaluate your formula for $f'(1/2)$ and find it is positive. For a quick way to do this without a calculator, note that $\pi/2<2<\pi$ so $\cos 2$ is negative, so the second term $6\cos(2a)$ will be positive. Since the last term $3\sin(2a)\sin(6a)$ cannot be less than $-3$, the entire expression for $f'(1/2)$ must be positive. So in that case (D) is correct.

Presumably the problem intended for you to assume that $f(x)$ is defined somewhere, so that (D) is the only correct answer. But this really should be stated explicitly if the problem is supposed to have a well-defined answer.