For the function \( f:\mathrm{R}\to\mathrm{R} \) given by \( f(x)=|x-3|+|x+4| \), where \( \mathrm{R} \) is set of real numbers, which of the following statements are correct?
(A) \( f(x) \) is continuous as well as differentiable at \( x=3 \)
(B) \( f(x) \) is continuous but not differentiable at \( x=3 \) and \( x=-4 \)
(C) \( f(x) \) is neither continuous nor differentiable at \( x=3 \)
(D) \( f(x) \) is continuous on \( \mathrm{R} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B) and (D) only
The modulus function is continuous everywhere, and a sum of continuous functions is continuous, so f is continuous on R. Differentiability can fail only where an expression inside a modulus is zero: x = 3 and x = −4.
Without modulus signs,
\( \displaystyle f(x)=\begin{cases}-2x-1, & x\lt -4\\7, & -4\le x\le 3\\2x+1, & x\gt 3\end{cases} \)
\( \displaystyle \text{At}\ x=3:\ Lf'(3)=0,\ Rf'(3)=2 \)
\( \displaystyle \text{At}\ x=-4:\ Lf'(-4)=-2,\ Rf'(-4)=0 \)
The one-sided derivatives differ at both points, so f is not differentiable at x = 3 or x = −4.
So f is continuous on R, including at x = 3 where f(3) = 7, and it is not differentiable at x = 3 and x = −4.
Checking the statements
(A) \( f(x) \) is continuous as well as differentiable at \( x=3 \) — Incorrect. \( Lf'(3)=0\ne 2=Rf'(3) \), so f is not differentiable at x = 3.
(B) \( f(x) \) is continuous but not differentiable at \( x=3 \) and \( x=-4 \) — Correct. The one-sided derivatives are 0 and 2 at x = 3, and −2 and 0 at x = −4.
(C) \( f(x) \) is neither continuous nor differentiable at \( x=3 \) — Incorrect. f is continuous at x = 3: both sides give f(3) = 7.
(D) \( f(x) \) is continuous on \( \mathrm{R} \) — Correct. A sum of modulus functions is continuous everywhere.