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Chapter 2: Problem 52

Find the local maximum and minimum values of the function and the value of \(x\) at which each occurs. State each answer correct to two decimal places. $$f(x)=3+x+x^{2}-x^{3}$$

Short Answer

Expert verified
Local max: 4 at \( x = 1 \); Local min: 3.37 at \( x = -0.33 \).

Step by step solution

01

Find the derivative

Find the first derivative of the function \( f(x) = 3 + x + x^2 - x^3 \) to determine the critical points. This will help in identifying where the function's slope is zero. The derivative is \( f'(x) = 1 + 2x - 3x^2 \).
02

Solve for critical points

Set the first derivative equal to zero to find the critical points: \[ 1 + 2x - 3x^2 = 0 \] Rearranging gives: \[ 3x^2 - 2x - 1 = 0 \] Solving this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -2 \), and \( c = -1 \). This gives: \( x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \).
03

Calculate the critical points

Calculate inside the square root: \[ 4 + 12 = 16 \] Thus: \[ x = \frac{2 \pm 4}{6} \] The critical points are \( x = 1 \) and \( x = -\frac{1}{3} \).
04

Find the second derivative

To determine the nature of each critical point (local maximum or minimum), find the second derivative \( f''(x) = 2 - 6x \).
05

Evaluate the second derivative at critical points

Calculate the second derivative at each critical point:For \( x = 1 \), \( f''(1) = 2 - 6 \times 1 = -4 \) (negative, indicating a local maximum).For \( x = -\frac{1}{3} \), \( f''\left(-\frac{1}{3}\right) = 2 - 6 \times \left(-\frac{1}{3}\right) = 4 \) (positive, indicating a local minimum).
06

Evaluate the function at critical points

Calculate the function's value at each critical point:For \( x = 1 \), \( f(1) = 3 + 1 + 1^2 - 1^3 = 4 \).For \( x = -\frac{1}{3} \), \( f\left(-\frac{1}{3}\right) = 3 - \frac{1}{3} + \left(-\frac{1}{3}\right)^2 - \left(-\frac{1}{3}\right)^3 \approx 3.37 \).

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Key Concepts

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

Derivative
The concept of the derivative is foundational in calculus. A derivative, in simple terms, measures how a function changes as its input changes. It tells us the slope of the function at any given point. By finding the derivative of a function, we can gain insights into the rate at which the function's value is changing.
  • The derivative of a function, typically denoted as \( f'(x) \), can be thought of as the instantaneous slope of the function \( f(x) \).
  • To find the derivative of a polynomial function, like our given function \( f(x) = 3 + x + x^2 - x^3 \), we differentiate each term separately.
  • The power rule is commonly used here, which says that the derivative of \( x^n \) is \( nx^{n-1} \).
Applying these rules:
  • The derivative of \( 3 \) (a constant) is 0.
  • The derivative of \( x \) is 1.
  • The derivative of \( x^2 \) is \( 2x \).
  • The derivative of \( -x^3 \) is \( -3x^2 \).
Thus, the derivative of the function is \( f'(x) = 1 + 2x - 3x^2 \). By studying this derivative, we can identify important features of the original function \( f(x) \).
Critical Points
Critical points occur where the derivative of a function is zero or undefined, and they are essential to find local extrema (maximums and minimums). To locate critical points:
  • Set the derivative equal to zero: \( f'(x) = 0 \).
  • Solve this equation to find points \( x \) where the slope of \( f(x) \) is zero. These are the potential points of maxima or minima.
For the given derivative \( 1 + 2x - 3x^2 \):
  • Set it equal to zero: \( 3x^2 - 2x - 1 = 0 \).
  • Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( x \).
  • Here, \( a = 3 \), \( b = -2 \), and \( c = -1 \), resulting in \( x = 1 \) and \( x = -\frac{1}{3} \).
These critical points are where the function changes direction, either reaching a peak (maximum) or pit (minimum).
Local Maximum and Minimum
Local maxima and minima are specific points in a function's graph where it reaches a peak or a trough relative to nearby points. Once the critical points are determined, we need to verify if they are maxima or minima.
  • Use the second derivative test, which involves calculating the second derivative \( f''(x) \).
  • The second derivative helps identify the nature of these critical points:
    • If \( f''(x) > 0 \), the function is concave up, indicating a local minimum.
    • If \( f''(x) < 0 \), the function is concave down, indicating a local maximum.
For our function, the second derivative \( f''(x) = 2 - 6x \):
  • At \( x = 1 \): \( f''(1) = -4 \), indicating a local maximum.
  • At \( x = -\frac{1}{3} \): \( f''(-\frac{1}{3}) = 4 \), indicating a local minimum.
Evaluate the function at these points:
  • Local maximum at \( x = 1 \) with \( f(1) = 4 \).
  • Local minimum at \( x = -\frac{1}{3} \) with \( f(-\frac{1}{3}) \approx 3.37 \).
These concepts help us understand not just where a function peaks or dips, but why it behaves that way around those points.

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