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Chapter 5: Problem 30

Give a complete graph of the function, and identify the location of all critical points and inflection poir s. Check your work with a graphing utility. $$4 x^{1 / 3}-x^{4 / 3}$$

Short Answer

Expert verified
Critical points at \( x = 0 \) and \( x = 1 \); inflection point at \( x = 2 \).

Step by step solution

01

Find the first derivative

To find critical points, we need the first derivative of the function. The function is given by \[ f(x) = 4x^{1/3} - x^{4/3}. \]The first derivative, using power rules, is\[ f'(x) = \frac{4}{3}x^{-2/3} - \frac{4}{3}x^{1/3}. \]
02

Solve for critical points

Critical points occur where the first derivative is zero or undefined. Set the first derivative to zero:\[ \frac{4}{3}x^{-2/3} - \frac{4}{3}x^{1/3} = 0. \]Factor out the common term:\[ \frac{4}{3}x^{-2/3}(1 - x) = 0. \]This gives two potential critical points: \( x = 0 \) and \( x = 1 \).
03

Find the second derivative

To find inflection points, we need the second derivative. The first derivative is:\[ f'(x) = \frac{4}{3}x^{-2/3} - \frac{4}{3}x^{1/3}. \] Differentiating again, we find\[ f''(x) = -\frac{8}{9}x^{-5/3} + \frac{4}{9}x^{-2/3}. \]
04

Solve for inflection points

Set the second derivative to zero to find potential inflection points:\[ -\frac{8}{9}x^{-5/3} + \frac{4}{9}x^{-2/3} = 0. \]Factor out the common term:\[ -\frac{4}{9}x^{-5/3}(2 - x) = 0. \]This implies that the second derivative changes sign at \( x = 2 \). Thus, the inflection point is \( x = 2 \).
05

Verify with a graphing utility

Using a graphing utility, plot the function \( f(x) = 4x^{1/3} - x^{4/3} \). Check if the graph shows critical points at \( x = 0 \) and \( x = 1 \), and an inflection point at \( x = 2 \). The graph should display these points accurately as confirmed by the graphing utility.

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Key Concepts

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

Critical Points
Critical points of a function are vital in determining where the function's graph has peaks and valleys. To find these, we focus on where the first derivative equals zero or is undefined. Here's a basic process to identify critical points:
  • Find the first derivative of the function.
  • Set this derivative equal to zero and solve for the variable.
  • Examine where the derivative is undefined, as these can also be critical points.
For the function given \[ f(x) = 4x^{1/3} - x^{4/3}, \]we found that the first derivative is\[ f'(x) = \frac{4}{3}x^{-2/3} - \frac{4}{3}x^{1/3}. \]By setting \( f'(x) = 0 \)and solving, we discovered that \( x = 0 \) and \( x = 1 \)are the critical points. These points are significant because they indicate where the function might change from increasing to decreasing or vice versa.
Inflection Points
Inflection points are where a curve changes concavity, from concave up to concave down, or the other way around. To determine where these points are, we use the second derivative.
  • Find the second derivative of the function.
  • Set this second derivative equal to zero and solve.
  • Check where the second derivative changes sign, confirming a change in concavity.
In our function, the second derivative was calculated as\[ f''(x) = -\frac{8}{9}x^{-5/3} + \frac{4}{9}x^{-2/3}. \]By setting \( f''(x) = 0 \),we determined that\( x = 2 \)is an inflection point. This point is essential for understanding the function's curvature, showing a transition in the graph's concave structure.
Graphing Utilities
Graphing utilities are powerful tools for checking the accuracy of analytical solutions. They allow us to visualize a function and inspect critical and inflection points on its graph effortlessly. Here鈥檚 how graphing utilities can aid our process:
  • Plot the function using a graphing tool.
  • Visually identify critical points where the slope of the tangent is zero (peaks and troughs).
  • Spot inflection points where the graph changes concavity.
For our function \( f(x) = 4x^{1/3} - x^{4/3}, \)a graph will show critical points at \( x = 0 \) and \( x = 1 \),as well as an inflection point at \( x = 2 \).Using a graphing utility confirms these points, providing a visual method of verification. This supportive approach helps cement your understanding of differential calculus concepts and their graphical representation.

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Most popular questions from this chapter

Give a complete graph of the rational function, and label the coordinates of the stationary points and inflection points. Show the horizontal and vertical asymptotes, and label them with their equations. Check your work with a graphing utility. $$\frac{2 x^{2}-1}{x^{2}}$$

(a) Find the limits of the function as \(x \rightarrow 0^{+}\) and \(x \rightarrow+\infty .\) (b) Give a complete graph of the function, and identify the location of all relative extrema and inflection points. Check your work with a graphing utility. $$\frac{\ln x}{x^{2}}$$

Let \(f(x)=\left(x^{3}+1\right) / x .\) Show that the graph of \(y=f(x)\) approaches the curve \(y=x^{2}\) "asymptotically" in the sense that \(\lim _{x \rightarrow+\infty}\left[f(x)-x^{2}\right]=0 \quad\) and \(\quad \lim _{x \rightarrow-\infty}\left[f(x)-x^{2}\right]=0\) Sketch the graph of \(y=f(x)\) showing this asymptotic behavior.

(a) Find the limits of the function as \(x \rightarrow 0^{+}\) and \(x \rightarrow+\infty .\) (b) Give a complete graph of the function, and identify the location of all relative extrema and inflection points. Check your work with a graphing utility. $$x \ln x$$

Find all values of \(x\) where the graph of $$y=\frac{2 x^{3}-3 x+4}{x^{2}}$$ crosses its oblique asymptote.
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