Finding the Maximum of a Continuous Function over a Closed Interval For function $f\left ( x \right )=4x^{3}-6x^{2},$ the maximum occurs in the interval $\left [ 1,2 \right ]$ when $x$ is equal to ___________

I got $x=0$ is maxima. Because, in that point $f''(x)<0$
But in answer, it is given though $f''(x)<0$ at $x=0,$ but at $x=2$ function $f(x)$  value maximum i.e. $8.$ But is it a correct procedure to get maximum.
Is maximum value and maxima are two different thing??
 A: Strong Hints: Identify the critical values of $f(x)$ in $[1,2]$, evaluate $f(x)$ at them, as well as $f(1)$ and $f(2)$ (as you have a closed interval)---the largest of these will be the absolute maximum of $f(x)$ over $[1,2]$. 
A: The maximum means the highest real value that the function spits out when you take the domain to be the interval provided. 
First take the derivative of $f(x)$.
$f'(x)=12x^2-12x=12x(x-1)$.
Now the second derivative is:
$f''(x)=24x-12$. The second derivative is clearly $>0$ in the interval $[1,2]$. 
So there occurs a MINIMA at the points where the first derivative is zero in the interval $[1,2]$, ie, when $x=1$. You won't consider and worry about $x=0$ as it is not in the interval. 
Now we know that $f(x)$ is a continuous function (why?), and the first derivative is always positive in the interval $[1,2]$. So it is strictly increasing in the interval.
Hence, the maximum value will be reached in the interval when $x=2$, that is the maximum value is $4(2)^3-6(2)^2= 8$.
A: You are told to find the maximum value of $f(x) = 4x^3 - 6x^2$ in the interval $[1,2]$.  So why did you simply assert $x = 0$ when this value is clearly not in the requested interval?
This is a common mistake students make:  the calculation goes flawlessly, but there is no understanding of what it means.
When searching for critical points satisfying $$\frac{df}{dx} = 12x^2 - 12x = 0,$$ we easily find $x(x-1) = 0$ or $x \in \{0, 1\}$.  Therefore, these are relative extrema of $f$.  By computing the second derivative, $$\frac{d^2f}{dx^2} = 24x - 12,$$ we find that at $x = 0$, $f''(0) < 0$, so the function is concave down, and at $x = 1$, $f''(1) > 0$, so the function is concave up.  All this tells us is that $x = 0$ is a relative maximum, and $x = 1$ is a relative minimum.
Since the only critical point of $f$ that lies in $[1,2]$ is a relative minimum, we now have to consider the value of $f$ at the endpoints of the interval.  At $x = 1$, $f(1) = -2$, and at $x = 2$, $f(2) = 8$.  Moreover, $f'(x) > 0$ whenever $x > 1$.  So we know that on $(1,2]$, $f$ is increasing, consequently the absolute maximum of $f$ on the interval $[1,2]$ occurs at $x = 2$ and equals $8$.
