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Chapter 4: Problem 43

The function \(f(x)\) is defined as under: \(f(x)=\left\\{\begin{array}{ll}3^{x} & :-1 \leq x \leq 1 \\ 4-x & : 1

Short Answer

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

Step by step solution

01

Checking Continuity at \(x=1\)

The function \(f(x)\) is defined as \(3^{x}\) for \(-1 \leq x \leq 1\) and as \(4-x\) for \(1 < x < 4\). To check if the function is continuous at \(x=1\), it is necessary to confirm the equality of the limit of the function as \(x\) approaches \(1\) from the left (which uses the function definition for \(-1 \leq x \leq 1\)) and the limt of the function as \(x\) approaches \(1\) from the right (which uses the function definition for \(1 < x < 4\)), and the function's value at \(x=1\).\n\nFirst, calculate the function's value at \(x=1\): \(f(1)=3^{1}=3\).\n\nNext, calculate the limit of \(f(x)\) as \(x\) approaches \(1\) from the left: \(\lim_{{x \to 1^-}}3^{x}=3^{1}=3\).\n\nFinally, calculate the limit of \(f(x)\) as \(x\) approaches \(1\) from the right: \(\lim_{{x \to 1^+}}(4-x)=4-1=3\).\n\nSince the function's value at \(x=1\) and the limits of the function as \(x\) approaches \(1\) from both directions are all equal to \(3\), the function is continuous at \(x=1\).
02

Checking Differentiability at \(x=1\)

The function \(f(x)\) is differentiable at a point if it is continuous at that point (which we have already established for \(x=1\)) and the limits of the derivative from both directions at that point are equal. Hence, we need to determine if the left-hand derivative and the right-hand derivative at \(x=1\) are the same. \n\nThe derivative of \(3^{x}\) with respect to \(x\) is \(3^{x}\ln(3)\), so the left-hand derivative at \(x=1\) is \(3^{1}\ln(3)=3\ln(3)\).\n\nThe derivative of \(4-x\) with respect to \(x\) is \(-1\), so the right-hand derivative at \(x=1\) is \(-1\).\n\nSince the left-hand derivative and the right-hand derivative at \(x=1\) are not equal (\(3\ln(3) \neq -1\)), the function is not differentiable at \(x=1\).

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Key Concepts

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

Piecewise Functions
A piecewise function is a type of function defined by multiple expressions based on different intervals of input. Think of it as a function with multiple rules, each one applying to a specific part of the domain. This is particularly useful when modeling scenarios that have different behaviors over different ranges.

For example, consider the function given in our problem:
  • For \(-1 \leq x \leq 1\), the function is defined by \(f(x) = 3^x\).
  • For \(1 < x < 4\), the function follows another rule: \(f(x) = 4-x\).
In essence, the function switches its ruling expression once you pass a certain value of \(x\). This method of definition is immensely practical when the situation or problem exhibits distinct characteristics across different segments of its domain. Understanding the defined intervals and applying the correct expression is key when working with piecewise functions.
Continuity at a Point
Continuity at a point means a function does not have any abrupt changes or jumps at that point. For a function \(f(x)\) to be continuous at \(x = c\),
  • The limit of \(f(x)\) as \(x\) approaches \(c\) from the left must equal the limit as \(x\) approaches \(c\) from the right.
  • Additionally, these limits must equal the function's value at that point, \(f(c)\).
In our exercise, investigating continuity at \(x = 1\), we found that:
  • The limit of the function as \(x\) approaches \(1\) from the left is \(3\).
  • The limit as \(x\) approaches \(1\) from the right is also \(3\).
  • The actual value of \(f(1)\) is \(3\).
Since there is no discrepancy among these values, the function is indeed continuous at \(x = 1\). This continuous behavior implies that you can graph the function and pass smoothly through this point without lifting your pencil from the paper.
Differentiability at a Point
Differentiability at a point refers to whether a function has a derivative at that point- meaning the function's rate of change is consistent and smooth at that spot. A function differentiable at \(x = c\) should not only be continuous at \(x = c\), but the derivatives from left and right must be equal as well.

In the given problem, we've examined the derivatives of \(f(x)\) from both directions at \(x = 1\):
  • For the segment \(3^x\), the left-hand derivative is \(3 \ln(3)\).
  • On the other hand, for the segment \(4-x\), the right-hand derivative is \(-1\).
The difference between these derivatives indicates there is a sharp corner or cusp at \(x = 1\). Hence, the function is not differentiable at this point, even though it is continuous. This scenario is typical in piecewise functions where different segments come together. Therefore, always check both the continuity and that the left and right derivatives are equal when assessing differentiability.

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Most popular questions from this chapter

The set of all the points where the function \(f(x)=\left\\{\begin{array}{ll}0 \quad & : x=0 \\ \frac{x}{1+e^{1 / x}} & : x \neq 0\end{array}\right.\), is differentiable is (a) \((0, \infty)\) (b) \((-\infty, \infty)-\\{0\\}\) (c) \((-\infty, 0)\) (d) \(R\).

Check the differentiability of the function \(f(x)=|x-2|\) at \(x=2\).

The number of points at which the function \(f(x)=\left|x-\frac{1}{2}\right|+|x-1|+\tan x\) does not have a deriva- tive in the interval \((0,2)\), (a) 1 (b) 2 (c) 3 (d) 4

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) \(-1\)

The value of \(k\) for which the function \(f(x)=\left\\{\begin{array}{ll}\sin (1 / x) & : x \neq 0 \\ k & ; x=0\end{array}\right.\) makes continuous at \(x=0\) is (a) 8 (b) 1 (c) \(-1\) (d) None
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