Problem 21 - Calculus of a Single Variable

Question

Applying the First Derivative Test In Exercises \(17-40\) , (a) find the critical numbers of \(f\) (if any), (b) find the open interval(s) on which the function is increasing or decreasing, (c) apply the First Derivative Test to identify all relative extrema, and (d) use a graphing utility to confirm your results. $$ f(x)=2 x^{3}+3 x^{2}-12 x $$

Step-by-Step Solution

Answer

The critical numbers of the function are -2 and 1. The function is decreasing on the intervals \((- \infty, 1)\) and increasing on the interval \((1, \infty)\). The function has a relative minimum at \(x = 1\), and does not have any relative maximum.

Step 1: Find the derivative

The derivative of \(f(x) = 2x^3 + 3x^2 -12x\) is found by applying the power rule of differentiation. The derivative would be \(f'(x) = 6x^2 + 6x - 12\).

Step 2: Find the critical numbers

Critical numbers are found by setting the derivative equal to zero and solving for \(x\). So, we have to solve \(6x^2 + 6x - 12 = 0\). Dividing the entire equation by 6 simplifies it to \(x^2 + x - 2 = 0\). This factors to \((x - 1)(x + 2) = 0\), which gives us the solutions \(x = 1\) and \(x = -2\). So 1 and -2 are the critical numbers.

Step 3: Find where the function is increasing or decreasing

Take test points in the intervals \(-\infty, -2\), \(-2, 1\), and \(1, \infty\) and plug them into the derivative. If the derivative is positive, then the function is increasing. If it’s negative, then it’s decreasing. For \(x < - 2\), try \(x = -3\), \(f'(-3) = -9\), which is less than \(0\), so the function is decreasing on \((- \infty, -2)\). For \(-2 < x < 1\), try \(x = 0\), \(f'(0) = -12\), which is less than \(0\), so the function is still decreasing on \((-2, 1)\). For \(x > 1\), try \(x = 2\), \(f'(2) = 12\), which is greater than \(0\), so the function is increasing on \((1, \infty)\).

Step 4: Apply the First Derivative Test

Since the function changes from decreasing to increasing at \(x = 1\), 1 is a relative minimum. The function does not change from decreasing to increasing or from increasing to decreasing at \(x = -2\), so -2 is not a relative extremum.

Step 5: Use a graphing utility

To confirm the results, plot the function using a graphing tool. The graph will show a relative minimum at \(x = 1\) and no relative maximum. Our results are confirmed.

Key Concepts

Critical Numbers

Critical numbers are important in calculus since they help us find where the behavior of a function might change. They occur when the derivative of a function equals zero or where the derivative does not exist. In simpler terms, these are the points where the slope of the tangent line is flat, or where there’s a possible "turn" in the curve.To find critical numbers, we start by computing the derivative of the given function. Here, using the power rule (which will be explained further in this article), we found the derivative of the function, given as: \[ f'(x) = 6x^2 + 6x - 12 \]We then set this derivative equal to zero, as critical points are found where the derivative is zero or undefined. Solving this equation gives us the points \( x = 1 \) and \( x = -2 \), our critical numbers in this scenario.

Increasing and Decreasing Intervals

The increasing and decreasing nature of a function helps us understand how the function behaves over specific intervals. A function is increasing when its derivative is greater than zero and decreasing when its derivative is less than zero.To determine these intervals:

• Pick test points in the intervals created by the critical numbers.
• Substitute these test points into the derivative.
• Check whether the derivative is positive or negative.

For the function given, the intervals we checked are \((-\infty, -2)\), \((-2, 1)\), and \((1, \infty)\):- In \((-\infty, -2)\), the derivative is negative, indicating the function decreases.- In \((-2, 1)\), the derivative is negative, so the function continues to decrease.- In \((1, \infty)\), the derivative is positive, indicating the function increases.

Relative Extrema

Relative extrema refer to the highest or lowest points within a particular interval of a function. These points are where the function changes its behavior from increasing to decreasing or vice versa.To find these extrema, we utilize the First Derivative Test:

• If a function changes from increasing to decreasing at a critical number, it’s a relative maximum.
• If it changes from decreasing to increasing, it's a relative minimum.

For our function, the derivative changes from negative to positive at \(x = 1\), indicating a relative minimum there. At \(x = -2\), the function does not change, so there is no extremum at this point.

Power Rule of Differentiation

The power rule of differentiation is a straightforward technique used to find the derivative of polynomial expressions of the form \(x^n\). It’s essential for simplifying derivatives of functions.The power rule states, for any real number \(n\),:\[ \frac{d}{dx} x^n = nx^{n-1} \]This means you multiply the exponent \(n\) by the variable's coefficient and decrease the exponent by one. For the function in the exercise, \(f(x) = 2x^3 + 3x^2 - 12x\), applying the power rule gives us:\[ f'(x) = 6x^2 + 6x - 12 \]Each term is differentiated separately, showing how convenient and effective the power rule is when dealing with polynomial functions.