Problem 8
Question
Use the definition of the derivative to show that the following functions are not differentiable at \(x=0\). \(f(x)=x^{4 / 5}\)
Step-by-Step Solution
Verified Answer
The function is not differentiable at \( x = 0 \) because the derivative at this point does not exist.
1Step 1: Recall the Definition of the Derivative
To determine the differentiability of a function at a point, we need to compute the derivative using the limit definition. For a function \( f(x) \), the derivative at \( x = c \) is given by: \[f'(c) = \lim_{{h \to 0}} \frac{{f(c+h) - f(c)}}{h}\]For this problem, we want to check whether \( f(x) = x^{4/5} \) is differentiable at \( x = 0 \).
2Step 2: Substitute in the Derivative Definition
Substituting the given function and the point \( x = 0 \) into the definition, we get: \[\lim_{{h \to 0}} \frac{{(0+h)^{4/5} - 0^{4/5}}}{h} = \lim_{{h \to 0}} \frac{{h^{4/5}}}{h} = \lim_{{h \to 0}} h^{-1/5}\]
3Step 3: Evaluate the Limit
Next, we need to evaluate the limit \( \lim_{{h \to 0}} h^{-1/5} \). As \( h \) approaches zero, \( h^{-1/5} \) approaches infinity. This indicates that the limit does not exist because it becomes unbounded as \( h \) gets closer to zero.
4Step 4: Conclude the Differentiability at the Point
Since the limit \( \lim_{{h \to 0}} h^{-1/5} \) does not exist, the derivative \( f'(0) \) also does not exist. Therefore, the function \( f(x) = x^{4/5} \) is not differentiable at \( x = 0 \).
Key Concepts
Function LimitsDerivative DefinitionNon-Differentiable Functions
Function Limits
When we talk about function limits, we are considering how a function behaves as the input approaches a particular point. In simpler words, it tells us what value the function is getting close to, as the input edge nearer to some value.
A limit is denoted by \( \lim_{{x \to c}} f(x) \), it represents the value that \( f(x) \) approaches as \( x \) gets arbitrarily close to \( c \).To better understand, imagine you're walking towards a finish line. The closer you get to the finish line, the more predictable it becomes where you will end up.
A limit is denoted by \( \lim_{{x \to c}} f(x) \), it represents the value that \( f(x) \) approaches as \( x \) gets arbitrarily close to \( c \).To better understand, imagine you're walking towards a finish line. The closer you get to the finish line, the more predictable it becomes where you will end up.
- Limits help us predict this finish without actually reaching it just yet.
- They deal with the trend, not the value at the point itself.
Derivative Definition
The derivative is a fundamental concept in calculus. It represents the rate at which a function is changing at any given point. Think of it as measuring how fast or slow something happens.
Using the definition, derivatives help us see how the value of a function changes as its input changes, symbolized as \( f'(x) \), the derivative of \( f \).Mathematically, the derivative at a point \( x = c \) can be found using the limit definition:
Using the definition, derivatives help us see how the value of a function changes as its input changes, symbolized as \( f'(x) \), the derivative of \( f \).Mathematically, the derivative at a point \( x = c \) can be found using the limit definition:
- \( f'(c) = \lim_{{h \to 0}} \frac{{f(c+h) - f(c)}}{h} \)
- If the limit results in a real number, the function is differentiable at that point.
- Conversely, if the limit doesn't exist or isn't finite, the function is not differentiable there.
Non-Differentiable Functions
Not all functions have nice, neat derivatives at every point. Some functions have sharp corners or vertical tangents where derivatives do not exist. These points of trouble indicate non-differentiability.
Knowing whether a function is non-differentiable at a point tells us about the function's geometry:
Knowing whether a function is non-differentiable at a point tells us about the function's geometry:
- If the limit for the derivative doesn't yield a real number, expect non-differentiability.
- Sharp edges or abrupt changes in direction lead to such behavior.
Other exercises in this chapter
Problem 7
Find the derivative of each function. $$ g(x)=\frac{1}{2} x^{4} $$
View solution Problem 8
Find functions \(f\) and \(g\) such that the given function is the composition \(f(g(x))\). $$ \sqrt{\frac{x-1}{x+1}} $$
View solution Problem 8
For each function, find: a. \(f^{\prime \prime}(x)\) and b. \(f^{\prime \prime}(3)\). $$ f(x)=\frac{x+2}{x} $$
View solution Problem 8
Find the derivative of each function. $$ f(x)=\frac{1}{3} x^{9} $$
View solution