/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 71 If a function \(f\) is continuou... [FREE SOLUTION] | 91Ó°ÊÓ

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If a function \(f\) is continuous at \(a\) and \(\lim _{x \rightarrow a}\left|f^{\prime}(x)\right|=\infty,\) then the curve \(y=f(x)\) has a vertical tangent line at \(a\) and the equation of the tangent line is \(x=a\) If a is an endpoint of a domain, then the appropriate one-sided derivative (Exercises \(67-68\) ) is used. Use this definition to answer the following questions. Verify that \(f(x)=x^{1 / 3}\) has a vertical tangent line at \(x=0\)

Short Answer

Expert verified
Question: Prove that the function \(f(x)=x^{1/3}\) has a vertical tangent line at \(x=0\). Answer: To prove that the function \(f(x)=x^{1/3}\) has a vertical tangent line at \(x=0\), we tested the continuity of the function at \(x=0\) and found that it was continuous. Then, we calculated the derivative of the function, \(f'(x) = \frac{1}{3} x^{-\frac{2}{3}}\), and evaluated the limit of the absolute value of the derivative as \(x\) approached \(0\). Since this limit was found to be \(\infty\), we can conclude that \(f(x)=x^{1/3}\) has a vertical tangent line at \(x=0\), with the equation of the tangent line being \(x=0\).

Step by step solution

01

Test Continuity of the Function at x=0

Continuity of a function \(f(x)\) at a point \(a\) is satisfied if 1. \(f(a)\) exists 2. \(\lim _{x \rightarrow a} f(x)\) exists 3. \(\lim _{x \rightarrow a} f(x) = f(a)\) For our function \(f(x) = x^{1/3}\), let us check the continuity at \(x=0\). We can see that \(f(0) = 0^{1/3}= 0\). So, \(f(a)\) exists. Next, let us calculate the limit as \(x \rightarrow 0\). We know that \(x^{1/3}\) is continuous for all \(x\). So we can directly calculate the limit: $$\lim _{x \rightarrow 0} x^{\frac{1}{3}} = 0^{\frac{1}{3}} = 0$$. As \(\lim _{x \rightarrow 0} f(x) = f(0)\), we can conclude that \(f(x)\) is continuous at \(x=0\).
02

Calculate the Derivative of the Function

We will find the derivative of \(f(x)=x^{1/3}\) by applying the power rule for differentiation: $$ f'(x) = \frac{d}{dx}(x^{1/3}) = \frac{1}{3} x^{-\frac{2}{3}} $$
03

Take Limit of the Absolute Value of the Derivative as x approaches a

We want to find the limit of the absolute value of the derivative of \(f(x) = x^{1/3}\) as \(x\) approaches \(0\). $$ \lim_{x \rightarrow 0} \left| \frac{1}{3} x^{-\frac{2}{3}}\right| $$ Since \(x^{-\frac{2}{3}} = \frac{1}{x^{\frac{2}{3}}}\), the expression inside the limit becomes: $$ \lim_{x \rightarrow 0} \left| \frac{1}{3x^{\frac{2}{3}}}\right| $$ As \(x\) approaches \(0\), the denominator \(3x^{\frac{2}{3}}\) approaches \(0\) which results in the fraction approaching infinity. Therefore, the limit is: $$ \lim_{x \rightarrow 0} \left| \frac{1}{3x^{\frac{2}{3}}}\right| = \infty $$
04

Conclusion

Since the function \(f(x)=x^{1/3}\) is continuous at \(x=0\), and \(\lim_{x \rightarrow 0}\left|f^{\prime}(x)\right| = \infty\), according to the definition, \(f(x)=x^{1/3}\) has a vertical tangent line at \(x=0\). The equation of the tangent line is \(x=0\).

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Key Concepts

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

Continuity of Functions
Understanding the continuity of functions is a fundamental aspect of calculus. For a function to be continuous at a point, it must satisfy three essential conditions:
  • The function value at that point, \(f(a)\), must exist.
  • The limit of the function as it approaches that point, \(\lim_{x \rightarrow a} f(x)\), must exist.
  • The function value and the limit at that point must be equal, meaning \(\lim_{x \rightarrow a} f(x) = f(a)\).
Let's see how these criteria apply to the function \(f(x) = x^{1/3}\) at \(x = 0\). At \(x = 0\), \(f(0) = 0^{1/3} = 0\), so the first condition is satisfied. Then we calculate the limit of the function as \(x\) approaches \(0\), giving us \(\lim_{x \rightarrow 0} x^{1/3} = 0\). Since the limit equals the function value, \(f(x)\) is continuous at \(x = 0\). This continuity sets the stage for investigating vertical tangents.
Derivative Calculation
Calculating the derivative of a function helps us understand its rate of change. For the function \(f(x) = x^{1/3}\), we apply the power rule for differentiation. The power rule states that if \(f(x) = x^n\), then \(f'(x) = n\cdot x^{n-1}\).
Using the power rule here gives us:\[f'(x) = \frac{d}{dx}(x^{1/3}) = \frac{1}{3} x^{-2/3}\]This derivative is a representation of the slope of the function at any given point other than \(x = 0\) since it involves a negative power of \(x\). Vertical tangents, however, require a deeper exploration of limits, which we will discuss next.
Limits in Calculus
Limits play a crucial role in understanding vertical tangents, as they allow us to examine the behavior of a function or its derivatives as they approach particular points. In this context, we are interested in the limit of the absolute value of the derivative of \(f(x) = x^{1/3}\) as \(x\) approaches \(0\):\[\lim_{x \rightarrow 0} \left| \frac{1}{3} x^{-2/3}\right|\]To simplify, think of \(x^{-2/3}\) as \(\frac{1}{x^{2/3}}\). Substituting this, the limit becomes:\[\lim_{x \rightarrow 0} \left| \frac{1}{3x^{2/3}}\right| \]As \(x\) approaches \(0\), the denominator \(3x^{2/3}\) gets smaller and smaller, causing the entire fraction to grow larger towards infinity. This result, \(\infty\), suggests an infinite slope or a vertical tangent line at \(x = 0\). Thus, the equation of the vertical tangent line is \(x = 0\). This use of limits to identify vertical tangents is a vital technique in calculus.

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