Problem 39 - Calculus with Concepts in Calculus

Question

Find the intervals on which the given Ir function is increasing and those on which it is decreasing. $$ g(x)=1 /(x+3) $$

Step-by-Step Solution

Answer

The function is decreasing on \((-\infty, -3)\cup(-3, \infty)\).

Step 1: Determine the Derivative

To find out where the function is increasing or decreasing, we first need to determine the derivative of the function. The derivative of the function \( g(x) = \frac{1}{x+3} \) can be found using the quotient rule or by rewriting it as \( g(x) = (x+3)^{-1} \). Using the power rule for derivatives, the derivative is \( g'(x) = -1(x+3)^{-2} = -\frac{1}{(x+3)^2} \).

Step 2: Set the Derivative Equal to Zero

Next, we set the derivative equal to zero to find the critical points. Since \( g'(x) = -\frac{1}{(x+3)^2} \), we note that this fraction is never zero because the numerator is a constant (\(-1\)) and the denominator is \((x+3)^2\) which is always positive for all \( x eq -3 \). Thus, there are no critical points due to \( g'(x) = 0 \).

Step 3: Analyze the Sign of the Derivative

Though there are no critical points, we evaluate the sign of \( g'(x) \) over its entire domain. Since \( g'(x) = -\frac{1}{(x+3)^2} \) is always negative (the negative sign and \((x+3)^2\) is positive wherever it is defined), \( g(x) \) is decreasing wherever \( g'(x) \) is defined.

Step 4: Determine the Domain of the Derivative

The function \( g(x) = \frac{1}{x+3} \) and its derivative \( g'(x) \) are undefined at \( x = -3 \). Therefore, the function \( g(x) \) is decreasing on all intervals where it is defined, specifically \((-\infty, -3)\cup(-3, \infty)\).

Key Concepts

Derivative

A derivative is a core concept in calculus that measures how a function changes as its input changes. It's like checking the speed of a car: it shows how fast or slow the car is going at any moment. In mathematical terms, the derivative of a function at a certain point quantifies the rate of change or the slope of the tangent line at that point.
In our example, to find the derivative of the function \( g(x) = \frac{1}{x+3} \), we first rewrite it as \( g(x) = (x+3)^{-1} \). This helps us apply the power rule more easily. Once we find the derivative as \( g'(x) = -\frac{1}{(x+3)^2} \), we can use it to analyze the function's behavior and understand on which intervals the function increases or decreases.

Increasing and Decreasing Intervals

To know when a function is moving upwards (increasing) or downwards (decreasing), we look at the sign of the derivative. If the derivative is positive over a region, the function is increasing in that region. Conversely, if the derivative is negative, the function is decreasing.
For our function, \( g'(x) = -\frac{1}{(x+3)^2} \) tells us a lot. Since this expression is always negative for all \( x eq -3 \), the function \( g(x) = \frac{1}{x+3} \) is decreasing everywhere it is defined. This occurs on the intervals \((- infty, -3)\cup(-3, \infty)\) because \( x = -3 \) is where the function becomes undefined, due to the division by zero making approaches from either side.

Quotient Rule

The quotient rule helps us find derivatives of functions that are ratios of two other functions, like \( \frac{f(x)}{g(x)} \). The rule states: if \( f(x) \) and \( g(x) \) are differentiable then:

• \( \left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \)

However, in our example, we chose to use the power rule instead after rewriting the function. This was possible because the denominator \( x+3 \) was easy to manage as a power of \( -1 \). Choosing the right rule often depends on which can simplify the work most easily.

Power Rule

The power rule is one of the simplest and most used tools for finding derivatives in calculus. When you have a function \( f(x) = x^n \) where \( n \) is any real number, the derivative is \( f'(x) = nx^{n-1} \).
In our task to differentiate \( g(x) = (x+3)^{-1} \), the power rule made our job simple. We took the exponent \( -1 \) and multiplied it by \( (x+3) \), then reduced the exponent by one to get \(-2\). The result was \( g'(x) = -1(x+3)^{-2} = -\frac{1}{(x+3)^2} \). Using the power rule was efficient and straightforward, especially for power expressions.