Problem 33 - Calculus Early Transcendentals

Question

Find the intervals on which \(f\) is increasing and decreasing. $$f(x)=-12 x^{5}+75 x^{4}-80 x^{3}$$

Step-by-Step Solution

Answer

The function \(f(x)\) is increasing on the intervals \((-\infty, 0)\), \((1, 4)\), and \((4, \infty)\) and decreasing on the interval \((0, 1)\).

Step 1: Find the derivative of the function

To find the critical points of the function, we need to differentiate it with respect to \(x\): $$f'(x) = \frac{d}{dx} (-12x^5 + 75x^4 - 80x^3) = -60x^{4}+300x^{3}-240x^{2}$$

Step 2: Find the critical points

To find the critical points, we need to set the derivative equal to zero: $$-60x^{4}+300x^{3}-240x^{2} = 0$$ We can factor out a common factor \(-60x^2\), to make the equation easier to solve: $$-60x^2(x^2 - 5x + 4) = 0$$ So, either \(-60x^2 = 0\) or \(x^2 - 5x + 4 = 0\). Solving the equation \(-60x^2 = 0\), the critical point is \(x = 0\). Solving the quadratic equation \(x^2 - 5x + 4 = 0\), we can factor it: $$(x - 1)(x - 4) = 0$$ So, the critical points are \(x = 1\) and \(x = 4\).

Step 3: Test the intervals

Now that we have our critical points \(x = 0\), \(x = 1\) and \(x = 4\), we can test the intervals in which the function is increasing or decreasing by checking the sign of the derivative in each interval: - Interval 1 (\(x < 0\)): Choose \(x = -1\): $$f'(-1) = -60(-1)^4 + 300(-1)^3 - 240(-1)^2 = -60 + 300 - 240 > 0$$ So, \(f(x)\) is increasing in this interval. - Interval 2 (\(0 < x < 1\)): Choose \(x = \frac{1}{2}\): $$f'\left(\frac{1}{2}\right) = -60\left(\frac{1}{2}\right)^4 + 300\left(\frac{1}{2}\right)^3 - 240\left(\frac{1}{2}\right)^2 = -\frac{15}{4} + \frac{75}{4} - 60 < 0$$ So, \(f(x)\) is decreasing in this interval. - Interval 3 (\(1 < x < 4\)): Choose \(x = 2\): $$f'(2) = -60(2)^4 + 300(2)^3 - 240(2)^2 = -1920 + 2400 - 960 > 0$$ So, \(f(x)\) is increasing in this interval. - Interval 4 (\(x > 4\)): Choose \(x = 5\): $$f'(5) = -60(5)^4 + 300(5)^3 - 240(5)^2 = -75000 + 375000 - 6000 > 0$$ So, \(f(x)\) is increasing in this interval.

Step 4: Conclusion

Based on our analysis of the intervals: - The function \(f(x)\) is increasing on the intervals \((-\infty, 0)\), \((1, 4)\), and \((4, \infty)\). - The function \(f(x)\) is decreasing on the interval \((0, 1)\).

Key Concepts

Derivative

When studying functions, the derivative is a key concept because it helps us understand how the function behaves at different points. The derivative of a function provides the rate at which the function's value is changing at any given point. In simpler terms, it's like measuring the slope of a graph at a specific point.

For the given function, \[f(x) = -12x^5 + 75x^4 - 80x^3\], we find the derivative \(f'(x)\) to analyze its behavior:\[f'(x) = -60x^4 + 300x^3 - 240x^2\].

Finding the derivative is integral for identifying the critical points, which are the points where the slope is zero or undefined. These critical points are potentially where the function shifts from increasing to decreasing or vice versa. By setting the derivative equal to zero, \[-60x^4 + 300x^3 - 240x^2 = 0\], we can solve for these critical values. It is at these values that the function can change its trend.

Increasing and Decreasing Functions

Understanding whether a function is increasing or decreasing on certain intervals involves checking the sign of its derivative. If the derivative is positive over an interval, the function is increasing there, and if the derivative is negative, the function is decreasing.

For the function \(f(x) = -12x^5 + 75x^4 - 80x^3\) with the derivative \[f'(x) = -60x^4 + 300x^3 - 240x^2\], we identified critical points \(x = 0, 1,\) and \(4\).

We then evaluated the derivative at various intervals determined by these critical points:

• For \(x < 0\), \(f'(x) > 0\), indicating \(f(x)\) is increasing.
• For \(0 < x < 1\), \(f'(x) < 0\), showing \(f(x)\) is decreasing.
• For \(1 < x < 4\), \(f'(x) > 0\), meaning \(f(x)\) is increasing again.
• For \(x > 4\), \(f'(x) > 0\), where \(f(x)\) continues to increase.

These evaluations illustrate the change in behavior of the function over different intervals.

Factorization of Polynomials

Factorization is a powerful tool used in finding critical points by simplifying polynomial equations. It involves expressing a polynomial as a product of its factors, which can make calculations easier and more intuitive.

For the derivative of the given function, \[f'(x) = -60x^4 + 300x^3 - 240x^2\], we factor out \(-60x^2\), resulting in:\[-60x^2(x^2 - 5x + 4) = 0\].

From here, we further factor \(x^2 - 5x + 4\) into \((x - 1)(x - 4)\), which simplifies finding the roots of the quadratic equation. Solving these factors gives us the critical points\(x = 0, 1,\) and \(4\).

Factorization decodes the polynomial's complexity, revealing simpler components to analyze critical points and other essential characteristics of the function.