Problem 30 - Thomas Calculus

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$ K(t)=15 t^{3}-t^{5} $$

Step-by-Step Solution

Answer

Function is increasing on \((-3,0)\) and \((0,3)\), decreasing on \((-\infty,-3)\) and \((3,\infty)\). Local min at \(t=-3\) is \(-162\); local max at \(t=3\) is \(162\).

Step 1: Find the derivative of the function

To determine the intervals of increase or decrease, we first find the derivative of the function. The derivative \( K'(t) \) of \( K(t) = 15t^3 - t^5 \) is: \[ K'(t) = \frac{d}{dt}(15t^3 - t^5) = 45t^2 - 5t^4 \]

Step 2: Solve for critical points

To find the critical points, solve \( K'(t) = 0 \): \[ 45t^2 - 5t^4 = 0 \] Factor out the common term: \[ 5t^2(9 - t^2) = 0 \] This gives: \[ 5t^2 = 0 \ \Rightarrow t = 0 \] or \[ 9 - t^2 = 0 \ \Rightarrow t = \pm 3 \] The critical points are \( t = 0, 3, -3 \).

Step 3: Determine intervals of increase and decrease

Use a sign chart for \( K'(t) \) around the critical points \( t = -3, 0, \) and \( 3 \):- Choose test points in the intervals: \((-\infty, -3), (-3, 0), (0, 3), (3, \infty)\)- For \( t = -4 \) in \((-\infty, -3)\): \( K'(-4) = 45(-4)^2 - 5(-4)^4 = 720 - 1280 < 0 \) (decreasing)- For \( t = -1 \) in \((-3, 0)\): \( K'(-1) = 45(-1)^2 - 5(-1)^4 = 40 > 0 \) (increasing)- For \( t = 1 \) in \((0, 3)\): \( K'(1) = 45(1)^2 - 5(1)^4 = 40 > 0 \) (increasing)- For \( t = 4 \) in \((3, \infty)\): \( K'(4) = 45(4)^2 - 5(4)^4 = 720 - 1280 < 0 \) (decreasing)Thus, \( K(t) \) is decreasing on \((-\infty, -3)\) and \((3, \infty)\), and increasing on \((-3, 0)\) and \((0, 3)\).

Step 4: Identify local and absolute extremes

Examine function values at critical points to identify local and absolute extrema:- At \( t = -3 \): \( K(-3) = 15(-3)^3 - (-3)^5 = -405 + 243 = -162 \), potential local minimum.- At \( t = 0 \): \( K(0) = 15(0)^3 - (0)^5 = 0 \) gives neither a minimum nor a maximum.- At \( t = 3 \): \( K(3) = 15(3)^3 - (3)^5 = 405 - 243 = 162 \), potential local maximum.Given the intervals of increase and decrease:- Local minimum at \( t = -3 \) with value \( -162 \).- Local maximum at \( t = 3 \) with value \( 162 \).- There is no absolute minimum or maximum as \( t \to \pm \infty \).

Step 5: Conclusion

The function \( K(t) = 15t^3 - t^5 \) is increasing on the intervals \((-3, 0)\) and \((0, 3)\), and decreasing on \((-\infty, -3)\) and \((3, \infty)\). It achieves a local minimum at \( t = -3 \) with value \(-162\) and a local maximum at \( t = 3 \) with value \( 162 \).

Key Concepts

Derivatives

Understanding derivatives helps us grasp how a function behaves. In simple terms, a derivative represents the rate of change of a function with respect to its variable. Think of it as the slope of the function at any given point. For the function \( K(t) = 15t^3 - t^5 \), the derivative is calculated to be \( K'(t) = 45t^2 - 5t^4 \). This tells us how the values of \( K(t) \) change as \( t \) changes.
When finding the derivative, you're essentially applying rules of differentiation that come down to power rules, product rules, and more, depending also on the complexity of the function. For polynomials like this one, using the power rule—differentiating terms of the form \( at^n \) by multiplying \( a \) by the exponent \( n \) and then decreasing the exponent by 1—is straightforward.

• The derivative helps identify critical points, where the function stops increasing or decreasing.
• Setting the derivative to zero enables us to solve for these critical points.

In our exercise, solving \( K'(t) = 0 \) provided us with the critical points \( t = 0, 3, \) and \( -3 \). Knowing this, we can further analyze how the function behaves around these crucial points.

Intervals of Increase and Decrease

The intervals on which a function is increasing or decreasing can highlight where the graph of the function rises or falls. Once we find the derivative and critical points from it, we can determine these intervals.
We use test points from the intervals formed by our critical points to check whether the derivative is positive or negative in each interval. If \( K'(t) > 0 \), the function is increasing; if \( K'(t) < 0 \), it is decreasing.

• For \( K(t) \), it's increasing on the intervals \((-3, 0)\) and \((0, 3)\).
• It's decreasing on \((-fty, -3)\) and \((3, fty)\).

This information guides us to understand how the function behaves as \( t \) varies, climbing upwards when increasing and descending when decreasing.

Local Extrema

Local extrema refer to the peaks and troughs of a function within a specific region, or simply put, a local maximum or minimum. At these points, the function switches its direction—peaking before descending or hitting a trough before rising again.
To identify local extrema for \( K(t) \), we consider the critical points found earlier and ascertain their nature based on surrounding intervals.

• At \( t = -3 \), the function has a local minimum since the function changes from decreasing to increasing.
• At \( t = 3 \), there's a local maximum as the function moves from increasing to decreasing.

By substituting these critical points back into \( K(t) \), we get the values:\[\begin{align*}\text{Local Minimum:} \ K(-3) &= -162,\text{Local Maximum:} \ K(3) &= 162.\end{align*}\]These do not serve as absolute extrema—the lowest or highest points overall—as the function continuously decreases and increases as \( t \) moves towards \( fty \). Therefore, for \( K(t) = 15t^3 - t^5 \), the local extrema are determined, informing us about the significant turning points in its course.