Question

At which point does f(x) = |x – 1| has itslocal minimum?

(a) They are unequal

(b) They are equal

(c) Depend on the numbers

(d) Can’t be predicted

The question was asked in examination.

My doubt stems from Calculus Application in section Application of Calculus of Mathematics – Class 12

Answer 1

Right answer is (b) They are equal

To explain I would say: The given function is f(x) = ∣x − 1∣, x ∈ R.

It is known that a function f is differentiable at point x = c in its domain if both

\(\lim\limits_{h \rightarrow 0^-}\) hf(c + h) – f(c)

And

\(\lim\limits_{h \rightarrow 0^+}\) hf(c + h) – f(c) are finite and equal.

To check the differentiability of the function at x = 1,

LHS,

Consider the left hand limit of f at x=1

\(\lim\limits_{h \rightarrow 0^-}\frac{|1+h-1|-|1-1|}{h}\)

= \(\lim\limits_{h \rightarrow 0^-}\frac{|h|}{h}\)

= \(\lim\limits_{h \rightarrow 0^-}\frac{-h}{h}\)

= −1

RHS,

Consider the right hand limit of f at x − 1

\(\lim\limits_{h \rightarrow 0^+}\frac{|1+h-1|-|1-1|}{h}\)

= \(\lim\limits_{h \rightarrow 0^+}\frac{|h|}{h}\)

= 1

Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1.

As, LHS = -1 and RHS = 1, it is clear that, f’(1) < 0 on the left of x = 1 and f’(x) > 0 on the right of the point x = 1.

Hence, f’(x) changes sign, from negative on the left to positive on the right of the point x = 1.

Therefore, f(x) has a local minima at x = 1.