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Chapter 4: Problem 44

A function is defined as follows: \(f(x)=\left\\{\begin{array}{ll}x^{2} & : x^{2}<1 \\ x & : x^{2} \geq 1\end{array}\right.\). The function is (a) continuous at \(x=1\) (b) differentiable at \(x=1\) (c) continuous but not differentiable at \(x=1\) (d) None.

Short Answer

Expert verified
(c) The function is continuous but not differentiable at $x=1$

Step by step solution

01

Understanding the function

The function provided is $f(x) = \left\{\begin{array}{ll} x^{2} & x^{2} < 1 \ x & x^{2} \geq 1 \end{array}\right.$. Note that the function indeed uses two different 'rules' or expressions depending on the value of $x$ as per the given conditions.
02

Checking the continuity of the function

A function is continuous at a specific point if the limit of the function as x approaches the point from the left equals the limit as x approaches the point from the right and equals the function value at the point. Let's calculate the limit for $x=1$ from both the left and the right as follows and compare it to the function value when $x=1$: \[ \lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^{2} = 1 \] \[ \lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}x = 1 \] \[ f(1) = 1 \] Therefore, $f(x)$ is continuous at $x=1$ as the left limit, the right limit, and the function value are all equal.
03

Checking the differentiability of the function

A function is differentiable at a point if it has a derivative at that point. That is, if the function doesn't exhibit any abrupt changes of direction (sharp corners/points) at the point. Now let's calculate the left and right derivatives at $x=1$. If they are equal, then the function is differentiable at $x=1$ based on the definition. But if not, then the function is not differentiable at that point. \[ \lim_{h\to 0^-} \frac{f(1+h)-f(1)}{h} = \lim_{h\to 0^-} \frac{((1+h)^2)-1}{h} = 2 \] \[ \lim_{h\to 0^+} \frac{f(1+h)-f(1)}{h} = \lim_{h\to 0^+} \frac{(1+h)-1}{h} = 1 \] Therefore, the left and right derivatives are not equal which means $f(x)$ is not differentiable at $x=1$ as it has a sharp corner or cusp.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuity of Functions
Continuity is a fundamental concept in differential calculus. A function is said to be continuous at a point if there are no breaks, jumps, or holes in its graph at that point.
In more formal terms, for a function \( f(x) \) to be continuous at a point \( x = a \), the following three conditions must be met:
  • The left-hand limit as \( x \to a^- \) must exist.
  • The right-hand limit as \( x \to a^+ \) must exist.
  • Both these limits should be equal and also equal to the function value \( f(a) \).
For the piecewise function in the exercise, we examined the value at \( x = 1 \) and found:
  • The limit from the left (using \( x^2 \) when \( x^2 < 1 \)) is 1.
  • The limit from the right (using \( x \) when \( x^2 \geq 1 \)) is also 1.
  • The function value \( f(1) \) is 1.
Since all these values are equal, the function is continuous at \( x = 1 \). This means there is no sudden jump at that point on the graph.
Differentiability of Functions
Differentiability refers to the ability of a function to have a derivative at every point in its domain. A function is differentiable at a point if it has a well-defined tangent at that point.
For a function to be differentiable at \( x = a \), the left-hand derivative and right-hand derivative must both exist and be equal. In practical terms:
  • Calculate the derivative from the left as \( x \to a^- \).
  • Calculate the derivative from the right as \( x \to a^+ \).
  • If these two derivatives are equal, the function is differentiable at that point.
In the given exercise, calculating at \( x = 1 \) showed:
  • The left derivative using \( x^2 \) is 2.
  • The right derivative using \( x \) is 1.
The inequality of these derivatives indicates that the function has a sharp corner at \( x = 1 \), making it not differentiable at that point. Differentiability is a stronger condition than continuity, so a function that is differentiable at a point will also be continuous there.
Piecewise Functions
Piecewise functions are functions that have different expressions based on specific intervals of the input \( x \). These types of functions are quite versatile as they allow a function to behave differently in different parts of its domain.
For the function in the exercise, it was defined as:
  • \( f(x) = x^2 \) when \( x^2 < 1 \).
  • \( f(x) = x \) when \( x^2 \geq 1 \).
Analyzing piecewise functions involves:
  • Identifying which expression is used based on the value of \( x \).
  • Calculating limits and derivatives from both sides of any discontinuity or potential corner point.
  • Checking conditions of continuity and differentiability at points where the function expression changes.
Piecewise functions can be continuous across different intervals, but the transition point where expressions change must be carefully analyzed for continuity and differentiability, as demonstrated in this exercise at \( x = 1 \). The change from one form to another can introduce discontinuities or non-differentiable corners.

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Most popular questions from this chapter

Check the differentiability of the function \(f(x)=\sin x+|\sin x|\) at \(x=0\).

Let \(f(x)= \begin{cases}\frac{x-4}{|x-4|}+a & : x<4 \\ a+b & : x=4 . \text { Then } f(x) \text { is con- } \\ \frac{x-4}{|x-4|}+b & : x>4\end{cases}\) tinuous at \(x=4\), when (a) \(a=b=0\) (b) \(a=b=1\) (c) \(a=-1, b=1\) (d) \(a=1, b=-1\).

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Given the function \(f(x)=\frac{1}{1-x} .\) The points of discontinuity of the composite function \(y=f(f(f(x))\) are at \(x=\) (a) 0 (b) 1 (c) 2 (d) \(-1\)

Find the number of the points of discontinuity of each of the following functions (i) \(f(x)=\frac{1}{x-1}\) (ii) \(f(x)=\frac{1}{|x|-1}\) (iii) \(f(x)=\frac{1}{|x|+2}\) (iv) \(f(x)=\frac{x^{2}+1}{x^{2}-4}\) (v) \(f(x)=\frac{1}{\log _{e} x}\) (vi) \(f(x)=\frac{1}{\log _{e}|x|}\) (vii) \(f(x)=\frac{1}{\log _{e}|x-2|}\) (viii) \(f(x)=\frac{1}{\log _{e}\left(x^{2}-1\right)}\) (ix) \(f(x)=\frac{1}{\log _{e}\left|\left(x^{2}-1\right)\right|}\) (x) \(f(x)=\frac{1}{x^{2}-3|x|+2}\) (xi) \(f(x)=\frac{x^{2}+4 x+1}{2 \sin x-1}\) (xii) \(f(x)=\frac{x^{2}-7 x+2014}{\sqrt{2} \cos x-1}\)
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