Let f(x)=∣∣2x2+5∣∣x|−3|,x∈R. If m and n denote the number of points where f is not continuous and not differentiable respectively, then m+n is equal to :

Question

Let f(x)=∣∣2x2+5∣∣x|−3|,x∈R. If m and n denote the number of points where f is not continuous and not differentiable respectively, then m+n is equal to :
🧐 Not the exact question you are looking for?Go ask a question

Solution 1

The function f(x) = |2x^2 + 5|/|x - 3| is a rational function where the numerator is an absolute value function and the denominator is also an absolute value function.

Step 1: Identify the points of discontinuity A rational function is discontinuous at the points where the denominator is zero. So, Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study prob

Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solv

This problem has been solved

Similar Questions

Let f : [0, π] → R be defined byf (x) =(0 if x = 0,x sin 1x − 1x cos 1x if x̸ = 0.Is f continuous?

Verified Answer

Let f(x)=∣∣2x2+5∣∣x|−3|,x∈R. If m and n denote the number of points where f is not continuous and not differentiable respectively, then m+n is equal to :

Let f(x)=2x−x2,x∈R. If m and n are respectively the number of points at which the curves y=f(x) and y=f′(x) intersects the x-axis, then the value of m+n is

Let f(x)=(x+3)2(x−2)3,x∈[−4,4]. If M and m are the maximum and minimum values of f, respectively in [−4,4], then the value of M−m is :

Let p ∈ Q. Define f (x) = xp for x > 0. Prove that f is differentiable on (0, ∞) and thatf ′(x) = pxp−1 for all x > 0

1/3