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

Answer 1

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.