Problem 20 - College Algebra

Question

Determine the interval(s) on which the function is increasing and decreasing. $$f(x)=4(x+1)^{2}-5$$

Step-by-Step Solution

Answer

The function is decreasing on \((-\infty, -1)\) and increasing on \((-1, \infty)\).

Step 1: Identify the function

The function given is \( f(x) = 4(x+1)^2 - 5 \). First, we must recognize that this is a quadratic function, which generally has the form \( ax^2 + bx + c \). Here, it is presented in vertex form, \( a(x - h)^2 + k \), where \( a = 4 \), \( h = -1 \), and \( k = -5 \).

Step 2: Find the derivative of the function

To determine where the function is increasing or decreasing, we find the derivative of \( f(x) \). The derivative of \( f(x) = 4(x+1)^2 - 5 \) with respect to \( x \) is \( f'(x) = 8(x+1) \).

Step 3: Find critical points

To find critical points, set the derivative equal to zero: \( 8(x+1) = 0 \). Solving this equation gives \( x = -1 \). This is the potential point where the behavior of \( f(x) \) changes.

Step 4: Determine intervals of increase and decrease

Analyze the sign of \( f'(x) = 8(x+1) \) around the critical point. For \( x < -1 \), choose test point \( x = -2 \), then \( f'(-2) = 8(-2+1) = -8 \), which is negative. For \( x > -1 \), choose test point \( x = 0 \), then \( f'(0) = 8(0+1) = 8 \), which is positive.

Step 5: Interpret results

Given that \( f'(x) \) changes from negative to positive at \( x = -1 \), the function decreases on the interval \( (-\infty, -1) \) and increases on \( (-1, \infty) \).

Key Concepts

Vertex Form

The vertex form of a quadratic function is an excellent tool for easily identifying important features of the graph, like its vertex, axis of symmetry, and orientation. Here, the function is given as \( f(x) = 4(x+1)^2 - 5 \), which aligns with the vertex form \( f(x) = a(x-h)^2 + k \). By comparing, we know:

• \( a = 4 \) - This tells us that the parabola opens upwards because \( a > 0 \) and is stretched vertically since \( a eq 1 \).
• \( h = -1 \) - This is the x-coordinate of the vertex. It indicates the horizontal offset from the origin.
• \( k = -5 \) - This is the y-coordinate of the vertex, which shows the vertical shift.

So, the vertex of the parabola is the point \((-1, -5)\). The vertex form simplifies identifying where the graph peaks or dips and aligns with finding the solution to the problem of intervals of increasing and decreasing behavior.

Derivative

The derivative of a function gives us a powerful tool for understanding how the function behaves. It essentially tells us how the function's value changes as the input changes. For our function, \( f(x) = 4(x+1)^2 - 5 \), the derivative is calculated as follows:Identifying the derivative is our first step in analyzing the function's behavior:

• We apply the power rule to find the derivative: \( f'(x) = 8(x+1) \).
• This derivative represents the rate of change of the function. In simpler terms, \( f'(x) \) tells us whether the function is going up or down at any given point.

Understanding how to calculate and use derivatives is foundational for identifying critical points and determining where a function is increasing or decreasing.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These are the points where the function might change from increasing to decreasing or vice versa, often corresponding to peaks, troughs, or points of inflection.For the function \( f(x) = 4(x+1)^2 - 5 \), the derivative is \( f'(x) = 8(x+1) \). Finding where this derivative equals zero will highlight critical points:

• Set \( 8(x+1) = 0 \) which gives us \( x = -1 \).
• This means \( x = -1 \) is a critical point where the behavior of the function may change.

The critical point reveals essential information about the graph's turning behaviors, helping us determine the function's increasing and decreasing intervals.

Intervals of Increase and Decrease

Determining the intervals on which a function increases or decreases involves analyzing its derivative's sign on either side of critical points.To find these intervals for \( f(x) = 4(x+1)^2 - 5 \):

• The critical point is \( x = -1 \), identified by setting the derivative \( 8(x+1) \) to zero.
• For \( x < -1 \), testing a value like \( x = -2 \) into the derivative gives \( f'(-2) = -8 \), which is negative, indicating the function is decreasing in \((-\infty, -1)\).
• For \( x > -1 \), testing a value like \( x = 0 \) into the derivative gives \( f'(0) = 8 \), which is positive, indicating the function is increasing in \((-1, \infty)\).

Understanding these changes helps visualize the function's graph as dipping down to a minimum at the vertex before rising again.