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Chapter 4: Problem 19

Find: (a) the intervals on which f is increasing, (b) the intervals on which f is decreasing, (c) the open intervals on which f is concave up, (d) the open intervals on which f is concave down, and (e) the x-coordinates of all inflection points. \(f(x)=3 x^{4}-4 x^{3}\)

Short Answer

Expert verified
(a) Increasing on \((1, \infty)\); (b) Decreasing on \((-\infty, 1)\); (c) Concave up on \((-\infty, 0)\) and \((\frac{2}{3}, \infty)\); (d) Concave down on \((0, \frac{2}{3})\); (e) Inflection points at \(x = 0\) and \(x = \frac{2}{3}\).

Step by step solution

01

Find the first derivative

To determine where the function is increasing or decreasing, we first find the first derivative of the function. The function given is \( f(x) = 3x^4 - 4x^3 \). The first derivative \( f'(x) \) is calculated as follows: \[ f'(x) = \frac{d}{dx}(3x^4 - 4x^3) = 12x^3 - 12x^2. \]
02

Solve for critical points

To find critical points, set the first derivative to zero and solve for \( x \):\[ 12x^3 - 12x^2 = 0. \]Factor the equation:\[ 12x^2(x - 1) = 0. \]The solutions are \( x = 0 \) and \( x = 1 \). These are the critical points.
03

Determine intervals of increase and decrease

Use the critical points to determine intervals: \((-\infty, 0)\), \((0, 1)\), and \((1, \infty)\). Test a point from each interval in the first derivative:- For \(x = -1\) in \((-\infty, 0)\), \( f'(-1) = 12(-1)^3 - 12(-1)^2 = -24 \) (negative, therefore decreasing).- For \(x = 0.5\) in \((0, 1)\), \( f'(0.5) = 12(0.5)^3 - 12(0.5)^2 = -1.5 \) (negative, therefore decreasing).- For \(x = 2\) in \((1, \infty)\), \( f'(2) = 12(2)^3 - 12(2)^2 = 48 \) (positive, therefore increasing).
04

Find the second derivative

To find intervals of concavity and inflection points, calculate the second derivative \( f''(x) \):\[ f''(x) = \frac{d}{dx}(12x^3 - 12x^2) = 36x^2 - 24x. \]
05

Solve for possible points of inflection

Set the second derivative to zero and solve for \( x \):\[ 36x^2 - 24x = 0. \]Factor the expression:\[ 12x(3x - 2) = 0. \]The solutions are \( x = 0 \) and \( x = \frac{2}{3} \). These are potential inflection points.
06

Determine intervals of concavity

Use potential inflection points to determine intervals: \((-\infty, 0)\), \((0, \frac{2}{3})\), and \((\frac{2}{3}, \infty)\). Test a point from each interval in the second derivative:- For \(x = -1\) in \((-\infty, 0)\), \( f''(-1) = 36(-1)^2 - 24(-1) = 60 \) (positive, therefore concave up).- For \(x = 0.5\) in \((0, \frac{2}{3})\), \( f''(0.5) = 36(0.5)^2 - 24(0.5) = -6 \) (negative, therefore concave down).- For \(x = 1\) in \((\frac{2}{3}, \infty)\), \( f''(1) = 36(1)^2 - 24(1) = 12 \) (positive, therefore concave up).
07

Identify inflection points

A change in concavity indicates an inflection point. From our intervals, the inflection points occur at the solutions from Step 5. Therefore, \( x = 0 \) and \( x = \frac{2}{3} \) are the inflection points.

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Key Concepts

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

First Derivative
The first derivative of a function provides vital information about the function's rate of change and its behavior over its domain. It tells us whether the function is increasing or decreasing in different intervals.
The given function is \( f(x) = 3x^4 - 4x^3 \). To find its first derivative, we apply basic differentiation rules:
  • The derivative of \( x^n \) is \( nx^{n-1} \).
  • So, \( f'(x) = \frac{d}{dx}(3x^4) - \frac{d}{dx}(4x^3) = 12x^3 - 12x^2 \).
With the first derivative \( f'(x) = 12x^3 - 12x^2 \), we can determine where the function is increasing or decreasing by solving \( f'(x) = 0 \) to find critical points that separate these intervals. These solutions are essential as they mark potential switch points in the behavior of the function.
Critical Points
Critical points are where the first derivative is zero or undefined. They are critical because they could indicate local maxima, minima, or points of non-monotonicity in a function. For the function \( f(x) = 3x^4 - 4x^3 \), we calculated the first derivative as \( f'(x) = 12x^3 - 12x^2 \).
To find the critical points:
  • Set \( f'(x) = 0 \) to get \( 12x^2(x - 1) = 0 \).
  • Solving this gives \( x = 0 \) and \( x = 1 \).
These critical values divide the number line into intervals where we can test the behavior of our function by substituting values into the derivative. Understanding the nature of these intervals and points helps to find where the function increases or decreases.
Intervals of Concavity
Concavity indicates whether the graph of a function is "bending" upwards or downwards. Identifying these intervals is crucial for understanding the behavior and shape of the graphed function.
We determine concavity using the second derivative \( f''(x) \). For our function, we found the second derivative to be \( f''(x) = 36x^2 - 24x \).
To identify the intervals of concavity:
  • Set \( f''(x) = 0 \) giving us \( 12x(3x - 2) = 0 \), hence \( x = 0 \) and \( x = \frac{2}{3} \).
  • These solutions help partition the domain into intervals: \((-finity, 0)\), \((0, \frac{2}{3})\), and \((\frac{2}{3}, finity)\).
  • Evaluate \( f''(x) \) at points within each interval to determine concavity:
    • Positive \( f''(x) \): graph is concave up.
    • Negative \( f''(x) \): graph is concave down.
Essentially, the sign of the second derivative informs us about the function's bending nature: upwards or downwards.
Inflection Points
An inflection point is where a graph changes its concavity. These points are essential as they indicate a "pivot" in the graph's shape. We use the second derivative to find these shifts.
For our function \( f(x) = 3x^4 - 4x^3 \), the potential inflection points were calculated by solving \( f''(x) = 0 \). The second derivative was \( f''(x) = 36x^2 - 24x \):
  • Solving gives \( x = 0 \) and \( x = \frac{2}{3} \).
To confirm these are true inflection points, check the sign of \( f''(x) \) around these values:
  • If the concavity changes from positive to negative (or vice versa), the point is an inflection point.
  • In our case, both \( x = 0 \) and \( x = \frac{2}{3} \) showed a change in concavity, thus they are indeed inflection points.
Identifying inflection points helps in drawing accurate graphs and understanding the overall movement of the function.

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Most popular questions from this chapter

Let \(f(x)=x^{2 / 3}, a=-1,\) and \(b=8\) (a) Show that there is no point \(c\) in \((a, b)\) such that $$ f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} $$ (b) Explain why the result in part (a) does not contradict the Mean-Value Theorem.

Writing A toy rocket is launched into the air and falls to the ground after its fuel runs out. Describe the rocket's acceleration and when the rocket is speeding up or slowing down during its flight. Accompany your description with a sketch of a graph of the rocket's acceleration versus time.

Let \(f(x)=\tan x\) (a) Show that there is no point \(c\) in the interval \((0, \pi)\) such that \(f^{\prime}(c)=0,\) even though \(f(0)=f(\pi)=0\). (b) Explain why the result in part (a) does not contradict Rolle's Theorem.

One way of proving that \(f(x) \leq g(x)\) for all \(x\) in a given interval is to show that \(0 \leq g(x)-f(x)\) for all \(x\) in the interval; and one way of proving the latter inequality is to show that the absolute minimum value of \(g(x)-f(x)\) on the interval is nonnegative. Use this idea to prove the inequalities in these exercises. Prove that \(\cos x \geq 1-\left(x^{2} / 2\right)\) for all \(x\) in the interval \([0,2 \pi] .\)

(a) Use the Mean-Value Theorem to show that if \(f\) is differentiable on an interval, and if \(\left|f^{\prime}(x)\right| \leq M\) for all values of \(x\) in the interval, then $$ |f(x)-f(y)| \leq M|x-y| $$ for all values of \(x\) and \(y\) in the interval. (b) Use the result in part (a) to show that $$ |\sin x-\sin y| \leq|x-y| $$ for all real values of \(x\) and \(y .\)
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