Consider the function f(x)=min |x2 - 4| , |x2 - 1| , then the number of points where f(x) is non-differentiable is/are

Question

Consider the function f(x)=min |x2 - 4| , |x2 - 1| , then the number of points where f(x) is non-differentiable is/are (A) 0 (B) 7 (C) 6 (D) 4

Tardigrade Admin · Accepted Answer

Using the graph of y=|x2 - 4|,y=|x2 - 1| <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-rdh0q5uqefkh.png" /> Clearly, from the graph we can see f(x) is non-differentiable at 6 points.

Q. Consider the function $f (x) = min {∣ ∣ x^{2} - 4 ∣ ∣, ∣ ∣ x^{2} - 1 ∣ ∣}$ , then the number of points where $f (x)$ is non-differentiable is/are

$0$

$7$

$6$

$4$

Using the graph of $y = ∣ ∣ x^{2} - 4 ∣ ∣, y = ∣ ∣ x^{2} - 1 ∣ ∣$

Clearly, from the graph we can see $f (x)$ is non-differentiable at $6$ points.

Q. Consider the function f(x)=min{∣∣​x2−4∣∣​,∣∣​x2−1∣∣​} , then the number of points where f(x) is non-differentiable is/are

0

7

6

4

Using the graph of y=∣∣​x2−4∣∣​,y=∣∣​x2−1∣∣​ Clearly, from the graph we can see f(x) is non-differentiable at 6 points.

Q. Consider the function $f (x) = min {∣ ∣ x^{2} - 4 ∣ ∣, ∣ ∣ x^{2} - 1 ∣ ∣}$ , then the number of points where $f (x)$ is non-differentiable is/are

$0$

$7$

$6$

$4$

Using the graph of $y = ∣ ∣ x^{2} - 4 ∣ ∣, y = ∣ ∣ x^{2} - 1 ∣ ∣$

Clearly, from the graph we can see $f (x)$ is non-differentiable at $6$ points.