Explanation Given Information: The function is f ( x ) = { 4 ( x − 5 ) 2 if x ≠ 5 − 3 if x = 5 . Formula used: Continuity of closed form function theorem: Every function that is closed is continuous on its domain If lim x → a f ( x ) exists and f ( a ) is defined, then lim x → a f ( x ) = f ( a ) The function is discontinuous at point x = a , if either, a) The limit lim x → a f ( x ) does not exist or b) The limit lim x → a f ( x ) does exist but it’s not equal to f ( a ) . Where, ‘ a ’ is any real number. Calculation: Consider the function f ( x ) = { 4 ( x − 5 ) 2 if x ≠ 5 − 3 if x = 5 At x = 5 , the function changes from one closed-form to another. So, the point x = 5 is the only possible point for discontinuity or a singular point. Check at x = 5 if the function is defined or not. At x = 5 the function is equal to − 3 , which is defined. Hence, it cannot be a singular point

7th Edition

ISBN: 9781337291248

Author: Waner, Stefan.

Publisher: Cengage Learning,

See similar textbooks

Concept explainers

Minimization

In mathematics, traditional optimization problems are typically expressed in terms of minimization. When we talk about minimizing or maximizing a function, we refer to the maximum and minimum possible values of that function. This can be expressed in ter…

Maxima and Minima

The extreme points of a function are the maximum and the minimum points of the function. A maximum is attained when the function takes the maximum value and a minimum is attained when the function takes the minimum value.

Derivatives

A derivative means a change. Geometrically it can be represented as a line with some steepness. Imagine climbing a mountain which is very steep and 500 meters high. Is it easier to climb? Definitely not! Suppose walking on the road for 500 meters. Which …

Concavity

In calculus, concavity is a descriptor of mathematics that tells about the shape of the graph. It is the parameter that helps to estimate the maximum and minimum value of any of the functions and the concave nature using the graphical method. We use the …

Videos

Question

Chapter 3.3, Problem 77E

To determine

To calculate: The singular points and points of discontinuity of the function f(x)={4(x−5)2 if x≠5−3 if x=5.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 3.3, Problem 76E

Chapter 3.3, Problem 78E

Students have asked these similar questions

4. Given the following information determine the appropriate trial solution to find yp. Do not solve the differential equation. Do not find the constants. a) (D-4)2(D+ 2)y = 4e-2x b) (D+ 1)(D² + 10D +34)y = 2e-5x cos 3x

3. Determine the appropriate annihilator for the given F(x). a) F(x) = 5 cos 2x b) F(x)=9x2e3x

Tangent planes Find an equation of the plane tangent to the following surfaces at the given points (two planes and two equations).

Chapter 3 Solutions

Applied Calculus

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

The singular points and points of discontinuity of the function f ( x ) = { 4 ( x − 5 ) 2 if x ≠ 5 − 3 if x = 5 .

Solution Summary: The author calculates the singular points and points of discontinuity of the function f(x)=l

Concept explainers

Videos

To calculate: The singular points and points of discontinuity of the function f(x)={4(x−5)2 if x≠5−3 if x=5.

Want to see the full answer?

Chapter 3 Solutions