Problem 34 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. $$ g(x)=x \sqrt{x+1} $$

Step-by-Step Solution

Answer

The function \(g(x) = x\sqrt{x+1}\) is decreasing in the interval \(x < -1\) and increasing in the intervals \(-1 < x <_\text{critical point}\) and \(x >_\text{critical point}\), where the critical point can be found by solving the equation \(0 = \sqrt{x+1} + \frac{x}{2}(x+1)^{-\frac{1}{2}}\).

Step 1: Apply the Product Rule to differentiate the given function

The given function is g(x) = x * √(x+1). Let's denote the two functions by u(x) = x and v(x) = √(x+1). Apply the Product Rule to find the derivative. The Product Rule states that (uv)' = u'v + uv'. Now, differentiate the two functions u(x) and v(x) with respect to x: \( u'(x) = \frac{d}{dx}(x) = 1 \\ v'(x) = \frac{d}{dx}(\sqrt{x+1}) = \frac{d}{dx}((x+1)^{\frac{1}{2}}) = \frac{1}{2}(x+1)^{-\frac{1}{2}} \) Now, apply the Product Rule: \(g'(x) = u'(x)v(x) + u(x)v'(x) = 1 * \sqrt{x+1} + x(\frac{1}{2}(x+1)^{-\frac{1}{2}}) \)

Step 2: Simplify the derivative

Now, simplify the derivative: \(g'(x) = \sqrt{x+1} + \frac{x}{2}(x+1)^{-\frac{1}{2}}\)

Step 3: Find the critical points

Set the derivative equal to zero to find the critical points: \(0 = \sqrt{x+1} + \frac{x}{2}(x+1)^{-\frac{1}{2}}\) The critical points are the values of x at which the function switches from increasing to decreasing or vice versa.

Step 4: Test intervals

Make a table to test intervals around the critical points by substituting test points in the derivative to see if the function is increasing or decreasing: | Interval | Test Point | Sign of \(g'(x)\) | Conclusion | |-------------------|--------------|-------------------|----------------------| | \(x < -1\) | -2 | - | Decreasing | | \(-1 < x <_\text{critical point}\)| 0 | + | Increasing | | \(x >_\text{critical point}\) | 1 | + | Increasing |

Step 5: State the intervals of increasing and decreasing

From the table above, the function g(x) is decreasing in interval \(x < -1\). Also, the function is increasing in interval \(-1 < x <_\text{critical point}\) and \(x >_\text{critical point}\). You can now substitute the actual critical point value into the results once you find it using a numerical method.

Key Concepts

Product Rule Differentiation

Understanding the product rule for differentiation is essential when analyzing functions that are the product of two or more sub-functions. In the case of the function g(x) = x \times \(\sqrt{x+1}\), we are dealing with just such a product: one function is u(x) = x and the other is v(x) = \(\sqrt{x+1}\).

The product rule states that the derivative of a product u(x)v(x) is given by u'(x)v(x) + u(x)v'(x). Applying this rule, a student can find the derivative of g(x) systematically by first finding the derivatives of u and v separately. For u(x) = x, the derivative u'(x) is 1. For v(x) = \(\sqrt{x+1}\), a transformation to (x+1)^{1/2} helps in applying the chain rule to find v'(x), yielding \frac{1}{2}(x+1)^{-1/2}\.

Combining these derivatives according to the product rule gives us the derivative of g(x), which is essential in analyzing the function's behavior.

Critical Points Calculus

In calculus, critical points are where a function's derivative is zero or undefined. Finding these points is a crucial step in function analysis as they indicate potential local maximums, minimums, or points of inflection. In our exercise, we find the critical points of g(x) by setting its derivative g'(x) to zero.

The equation \(0 = \sqrt{x+1} + \frac{x}{2}(x+1)^{-1/2}\) represents the condition for critical points. Solving this equation may sometimes require numerical methods, but it's the conceptual understanding that the critical points indicate where the function changes its behavior that guides further analysis. The derivative test involves picking values (test points) from intervals around these critical points to determine the function's increasing and decreasing nature.

Function Analysis

The heart of function analysis involves studying the way functions increase or decrease, and their overall behavior. After obtaining the derivative of g(x) and finding its critical points, a student can map out where the function is increasing or decreasing.

This is done by testing intervals around the critical points: selecting numbers (test points) within the intervals and plugging them into the derivative. The sign of the derivative at these points indicates whether the function is increasing (positive derivative) or decreasing (negative derivative). So in this exercise, by testing intervals and examining the signs of g'(x), it's concluded that g(x) is decreasing for \(x < -1\) and increasing for \(x > -1\), aside from the precise critical point where its behavior changes.