 QUESTION

# If the value of f(x) = |x – 2012|, then f’(x) at x = 2011 is given by (a) 1 (b) -1 (c) 0 (d) 2011

Hint: Here, we will first form a piecewise function and then check that in which interval the value of x lies. Then we will use the value of the function at that point to get the required value.

Complete Step-by-Step solution:
A piecewise- defined function is a function defined by multiple sub-functions, a sub-domain. Piecewise is actually a way of expressing the function rather than a characteristic of the function itself, but with additional qualification, it can describe the nature of the function. The word piecewise is also used to describe any property of a piecewise-defined function that holds for each piece but not necessarily holds for the whole domain of the function.
The modulus function returns a positive value of a variable or an expression. If reference to modulus of an independent variable, the function results in a non-negative value of the variable, irrespective of whether the independent variable is positive or negative.
When the value of x is a non-negative number, then function returns x and if the value of x is a negative number, then the function returns –x.
Since, the function given to us is:
$f\left( x \right)=|x-2012|$
We can write this function as piecewise function as:
f\left( x \right)=\left\{ \begin{align} & x-2012;x>2012 \\ & -\left( x-2012 \right);x<2012 \\ & 0;x=2012 \\ \end{align} \right.
So, when x =2011, we have:
\begin{align} & f\left( x \right)=-\left( x-2012 \right) \\ & \Rightarrow f\left( x \right)=-x+2012 \\ \end{align}
On differentiating it, we get:
$f'\left( x \right)=-1+0=-1$
Therefore, f’(x) at x = 2011 is -1.
Hence, option (b) is the correct answer.

Note: Students should be careful while writing f(x) as a piecewise function. This sign of the value returned by f(x) must be kept in mind.