Problem 2 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, use a table similar to that in Example 1 to find all relative extrema of the function. $$ f(x)=x^{2}+8 x+10 $$

Step-by-Step Solution

Answer

The function \(f(x)=x^{2}+8 x+10\) has a relative minimum at the point \(x=-4\).

Step 1: Find the derivative

Find the derivative of the function \(f(x)\). The derivative of the function is given by \(f'(x)=2x+8\).

Step 2: Identify Critical Points

Critical points are the points where the derivative either equals zero or does not exist. To find critical points, set the derivative equal to zero and solve for \(x\). This gives \(2x+8=0\), which simplifies to \(x=-4\). So, the function has one critical point at \(x=-4\).

Step 3: Performing the first derivative test

Plot the critical point on a number line, and pick test values from the intervals that they divide the number line into. These test values will be plugged back into the derivative function to assess where the function is increasing and decreasing. After this process, it will be clear that the function is decreasing on the interval \((-∞, -4)\) and increasing on the interval \((-4, ∞)\). So, the critical point at \(x=-4\) is a relative minimum.

Key Concepts

Understanding the Derivative

The derivative of a function captures how the function value changes as its input changes. Think of it as the "speed" at which the function is moving. When we differentiate a function, we are essentially finding its rate of change.

• For any given function, say \( f(x) = x^2 + 8x + 10 \), the derivative tells us how \( f(x) \) changes with a small increment in \( x \).
• The symbol for the derivative is typically \( f'(x) \). For our example function, the derivative is \( f'(x) = 2x + 8 \).

It's important to note that differentiating a polynomial like ours reduces its degree by one, which makes it an easier expression to handle when solving for critical points.

Finding Critical Points

Critical points are crucial places on a graph where the function's behavior changes. These occur where the derivative equals zero or is undefined. They help us locate places on the graph that might be peaks, troughs, or level areas.

• To find critical points for \( f(x) = x^2 + 8x + 10 \), we set the derivative \( f'(x) = 2x + 8 \) to zero and solve for \( x \).
• In our case, solving \( 2x + 8 = 0 \) gives \( x = -4 \), a single critical point.

The critical points let us know where to focus when determining the intervals of increase or decrease in a function. These intervals will later help identify relative minimums or maximums.

Using the First Derivative Test

The first derivative test helps us identify the nature of critical points—whether they are relative maxima, minima, or neither. By examining the sign of the derivative on intervals around the critical points, we can determine how the function behaves.

• For \( x = -4 \), plug test values from intervals around this critical point into \( f'(x) \).
• For values less than -4, such as -5, the derivative \( f'(x) = 2(-5) + 8 = -2 \) is negative, showing a decrease.
• For values greater than -4, like 0, the derivative \( f'(x) = 2(0) + 8 = 8 \) is positive, showing an increase.

The sign change from negative to positive at \( x = -4 \) indicates that it is a relative minimum. The function decreases to \( x = -4 \) and then increases, forming a "valley" at this point.