Problem 36 - Calculus Early Transcendentals

Question

Find the intervals on which \(f\) is increasing and decreasing. $$f(x)=\frac{x^{4}}{4}-\frac{8 x^{3}}{3}+\frac{15 x^{2}}{2}+8$$

Step-by-Step Solution

Answer

Answer: The function is increasing in the intervals (0, 3) and (5, ∞), and decreasing in the intervals (-∞, 0) and (3, 5).

Step 1: Calculate the derivative of the function f(x)

To find the derivative of the function, we will apply the power rule (\(\frac{d}{dx}(x^n) = nx^{n-1}\)): $$f'(x)=\frac{d}{dx}\left(\frac{x^{4}}{4}-\frac{8 x^{3}}{3}+\frac{15 x^{2}}{2}+8\right)$$ $$=4*\frac{1}{4}x^{4-1} - 3*\frac{8}{3}x^{3-1}+2*\frac{15}{2}x^{2-1}+0$$ $$=x^3 - 8x^2 + 15x$$

Step 2: Find the critical points

Critical points occur when the derivative, \(f'(x)\), is equal to zero or does not exist. In this case, the derivative is a polynomial function and is defined for all values of x. Thus, we only need to find where \(f'(x) = 0\): $$x^3 - 8x^2 + 15x = 0$$ Factoring x out, we get: $$x(x^2 - 8x + 15) = 0$$ Now we need to find the roots of the quadratic equation \(x^2 - 8x + 15\): $$x^2 - 8x + 15 = 0$$ Factoring it further, we get: $$(x-3)(x-5)=0$$ So, the critical points are x = 0, x = 3, and x = 5.

Step 3: Determine the intervals for increasing or decreasing

Using these critical points, divide the number line into intervals: \((-\infty, 0), (0, 3), (3, 5), (5, \infty)\). Choose a test point from each interval and plug it into the derivative, \(f'(x)\), to determine if the function is increasing (if \(f'(x) > 0\)) or decreasing (if \(f'(x) < 0\)) in that interval. Interval \((-\infty, 0)\): Test point \(x = -1\). $$f'(-1) = -1^3 - 8(-1)^2 + 15(-1) = -1 - 8 - 15 = -24 < 0$$ The function is decreasing in the interval \((-\infty, 0)\). Interval \((0, 3)\): Test point \(x=1\). $$f'(1) = 1^3 - 8(1)^2 + 15(1) = 1 - 8 + 15 = 8 > 0$$ The function is increasing in the interval \((0, 3)\). Interval \((3, 5)\): Test point \(x=4\). $$f'(4) = 4^3 - 8(4)^2 + 15(4) = 64 - 128 + 60 = -4 < 0$$ The function is decreasing in the interval \((3, 5)\). Interval \((5, \infty)\): Test point \(x=6\). $$f'(6) = 6^3 - 8(6)^2 + 15(6) = 216 - 288 + 90 = 18 > 0$$ The function is increasing in the interval \((5, \infty)\). In conclusion, the function \(f(x)\) is increasing in the intervals \((0,3)\) and \((5,\infty)\), and decreasing in the intervals \((-\infty, 0)\) and \((3, 5)\).

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes. They provide a tool to understand how a function behaves by indicating whether it is increasing or decreasing at any given point.
The derivative of a function, usually denoted by \(f'(x)\), is found using differentiation rules. For example, if you have a polynomial like \(f(x) = \frac{x^4}{4} - \frac{8x^3}{3} + \frac{15x^2}{2} + 8\), you apply the power rule to each term. The power rule states that the derivative of \(x^n\) is \(nx^{n-1}\). So for each term in our function, we differentiate:

• The derivative of \(\frac{x^4}{4}\) is \(x^3\).
• For \(-\frac{8x^3}{3}\), the derivative is \(-8x^2\).
• And \(\frac{15x^2}{2}\) becomes \(15x\).

The constant \(8\) drops off since it doesn't change with \(x\). Putting it all together, we get the derivative: \(f'(x) = x^3 - 8x^2 + 15x\).
Derivatives tell us the slopes of the tangent lines to the curve of the function, helping to identify how the function is rising or falling at specific intervals.

Identifying Critical Points

Critical points are specific values of \(x\) where a function’s derivative is zero or undefined. These points often signify a change from increasing to decreasing behavior (or vice versa). For the function \(f(x)\) given earlier, the derivative \(f'(x) = x^3 - 8x^2 + 15x\) is a polynomial, which is defined for all real numbers. So, we only need to find where \(f'(x) = 0\).
Setting \(f'(x) = 0\), we solve \(x^3 - 8x^2 + 15x = 0\) by factoring out \(x\), resulting in \(x(x^2 - 8x + 15) = 0\). This gives us one root at \(x = 0\). Further factorizing \(x^2 - 8x + 15\) gives \((x-3)(x-5)\), yielding two more roots: \(x = 3\) and \(x = 5\).
In summary, the critical points of the function are where \(x = 0, 3,\) and \(5\). These values are crucial for determining when a function might change from increasing to decreasing or vice versa.

Intervals of Increase and Decrease

Finding intervals where the function increases or decreases involves testing the derivative \(f'(x)\) with selected test points. These intervals are divided by the critical points identified previously.
For each interval derived from \(x = 0, 3, 5\), plug a test point into \(f'(x)\):

• For the interval \((-fty, 0)\), if \(x = -1\), then \(f'(-1) = -24 < 0\), indicating the function is decreasing.
• In the interval \((0, 3)\) with \(x = 1\), \(f'(1) = 8 > 0\), so the function is increasing.
• For \((3, 5)\), try \(x = 4\) with \(f'(4) = -4 < 0\), signaling a decrease.
• In \((5, fty)\) using \(x = 6\), \(f'(6) = 18 > 0\), the function increases.

The pattern of increasing and decreasing behavior forms the function's intervals; \((0, 3)\) and \((5, fty)\) are where it increases, while it decreases from \((-fty, 0)\) and \((3, 5)\). With these intervals identified, we gain insight into how the function behaves across its entire domain.