Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 5-26, identify the critical points and find the maximum value and minimum value on the given interval. $$ G(x)=\frac{1}{5}\left(2 x^{3}+3 x^{2}-12 x\right) ; I=[-3,3] $$

Step-by-Step Solution

Answer

Max: 9 at \(x = 3\); Min: \(-\frac{7}{5}\) at \(x = 1\).

Step 1: Find the derivative of G(x)

To find the critical points, we first need the derivative of the function. Differentiate \(G(x) = \frac{1}{5}(2x^3 + 3x^2 - 12x)\). Applying the power rule, the derivative is \(G'(x) = \frac{1}{5}(6x^2 + 6x - 12)\), simplifying to \(G'(x) = \frac{1}{5}(6(x^2 + x - 2))\). We can further simplify to \(G'(x) = \frac{6}{5}(x^2 + x - 2)\).

Step 2: Solve G'(x) = 0 to find critical points

We set the derivative \(G'(x) = \frac{6}{5}(x^2 + x - 2)\) equal to zero to find critical points. Simplifying, we get \(x^2 + x - 2 = 0\). Factor this quadratic equation to get \((x - 1)(x + 2) = 0\). Solving each factor gives critical points at \(x = 1\) and \(x = -2\).

Step 3: Evaluate G(x) at critical points and endpoints of the interval

Substitute \(x = 1\), \(x = -2\), and the endpoints of the interval \(x = -3\) and \(x = 3\) into \(G(x)\) to evaluate the function values:1. \(G(1) = \frac{1}{5}(2(1)^3 + 3(1)^2 - 12(1)) = \frac{1}{5}(-7) = -\frac{7}{5}\).2. \(G(-2) = \frac{1}{5}(2(-2)^3 + 3(-2)^2 - 12(-2)) = \frac{1}{5}(-8 + 12 + 24) = \frac{28}{5}\).3. \(G(-3) = \frac{1}{5}(2(-3)^3 + 3(-3)^2 - 12(-3)) = \frac{1}{5}(-54 + 27 + 36) = \frac{9}{5}\).4. \(G(3) = \frac{1}{5}(2(3)^3 + 3(3)^2 - 12(3)) = \frac{1}{5}(54 + 27 - 36) = \frac{45}{5} = 9\).

Step 4: Compare each G(x) value to determine max and min

The function values are \(G(1) = -\frac{7}{5}\), \(G(-2) = \frac{28}{5}\), \(G(-3) = \frac{9}{5}\), and \(G(3) = 9\). The maximum value within the interval is \(9\) at \(x = 3\), and the minimum value is \(-\frac{7}{5}\) at \(x = 1\).

Key Concepts

Derivative Calculation

Understanding how to find the derivative of a function is crucial for identifying critical points. In this exercise, we start with the function \[ G(x) = \frac{1}{5}(2x^3 + 3x^2 - 12x) \]. By applying the power rule, which states that if \( f(x) = ax^n \), then \( f'(x) = nax^{n-1} \), we can differentiate this polynomial function step-by-step.
The derivative calculation involves taking each term of the polynomial:

• For \( 2x^3 \), the derivative is \( 6x^2 \).
• For \( 3x^2 \), it becomes \( 6x \).
• For the \(-12x \) term, the derivative is \(-12 \).

Putting it together, the derivative \( G'(x) \) is given by \[ G'(x) = \frac{1}{5}(6x^2 + 6x - 12) \]. Simplifying further, we have \( G'(x) = \frac{6}{5}(x^2 + x - 2) \). The simplified form makes it easier for solving the quadratic equation in the next stages.

Quadratic Equation Solving

After calculating the derivative, the next step is to solve \( G'(x) = 0 \) to find the critical points. This means we will solve the quadratic equation \[ \frac{6}{5}(x^2 + x - 2) = 0 \].
It's often useful to simplify this by ignoring constant factors that don't affect the solution for \( x \). Thus, solving \[ x^2 + x - 2 = 0 \] is enough. Quadratic equations can usually be factored, but when they can't, the quadratic formula can be applied:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
For our specific equation, however, we can factor directly to:\[ (x - 1)(x + 2) = 0 \].
Solving these factors gives the critical points: \( x = 1 \) and \( x = -2 \). These solutions provide potential locations for maximum or minimum values or points of inflection on the graph of the function.

Endpoint Evaluation

Evaluating the function at the endpoints of the given interval and at the critical points ensures that we identify the absolute maximum and minimum values entirely. The endpoints in this exercise are \(-3\) and \(3\).
By substituting these values into \( G(x) \), we calculate:

• At \( x = -3 \), \( G(-3) \) gives us \( \frac{9}{5} \).
• At \( x = 3 \), \( G(3) \) becomes \( 9 \).

Evaluating the function at the critical points also yields:

• \( G(1) = -\frac{7}{5} \)
• \( G(-2) = \frac{28}{5} \)

These values represent possible extremes over the interval \([-3,3]\). It's essential to not only consider the critical points but also the endpoints because the maximum or minimum of the entire interval might occur there instead.

Interval Analysis

Expanding on evaluation, interval analysis involves examining both the endpoint values and the values at critical points to determine the overall behavior of the function within the given interval. For this function, knowing how the values compare helps confirm the global maximum and minimum.
We have calculated the function values and simplifying and comparing them shows:

• Highest value, or maximum, is at \( x = 3 \) with \( G(3) = 9 \).
• Lowest value, or minimum, is at \( x = 1 \) with \( G(1) = -\frac{7}{5} \).

Interval analysis helps you to cement the understanding that while critical points give important local information, the endpoints are just as significant when considering defined intervals. This thorough analysis ensures that nothing is overlooked in determining the absolute extrema across the given range.