Problem 16 - Calculus Early Transcendentals

Question

Find: (a) the intervals on which f is increasing, (b) the intervals on which f is decreasing, (c) the open intervals on which f is concave up, (d) the open intervals on which f is concave down, and (e) the x-coordinates of all inflection points. \(f(x)=5-4 x-x^{2}\)

Step-by-Step Solution

Answer

(a) Increasing on \(( -\infty, -2)\); (b) Decreasing on \((-2, \infty)\); (c) No concave up intervals; (d) Concave down on \(( -\infty, \infty)\); (e) No inflection points.

Step 1: Find the first derivative

First, we need to find the derivative of the function to determine the intervals where the function is increasing or decreasing. The given function is \( f(x) = 5 - 4x - x^2 \). Differentiating with respect to \( x \), we get:\[ f'(x) = -4 - 2x \]

Step 2: Find critical points

Critical points occur where the first derivative is zero or undefined. The equation from Step 1 is \( f'(x) = -4 - 2x = 0 \). Solving for \( x \), we get:\[ -4 - 2x = 0 \2x = -4 \x = -2 \]

Step 3: Test intervals for increasing/decreasing

We have a critical point at \( x = -2 \). We'll test the intervals \(( -\infty, -2)\) and \((-2, \infty)\) to see where the function is increasing or decreasing:- For \( x = -3 \) (in \( ( -\infty, -2) \)), \( f'(-3) = -4 - 2(-3) = 2 \) (positive, increasing)- For \( x = 0 \) (in \( (-2, \infty) \)), \( f'(0) = -4 \) (negative, decreasing)Thus, \( f(x) \) is increasing on \(( -\infty, -2)\) and decreasing on \((-2, \infty)\).

Step 4: Find the second derivative

To determine concavity, we find the second derivative of the function. From \( f(x) = 5 - 4x - x^2 \), the first derivative is \( f'(x) = -4 - 2x \). Differentiating again gives:\[ f''(x) = -2 \]

Step 5: Determine intervals of concavity

The second derivative is \( f''(x) = -2 \), which is constant and negative. Therefore, the function is concave down on the entire real line \( ( -\infty, \, \infty ) \). Since \( f''(x) = -2 \) does not change sign, there are no intervals of concavity changes, nor any concave up intervals.

Step 6: Identify inflection points

Inflection points occur where there is a change in concavity, i.e., where the second derivative changes sign. However, \( f''(x) = -2 \) is constant and does not change sign. Thus, there are no inflection points.

Key Concepts

Derivatives

In calculus, the derivative of a function at a point essentially tells us the slope of the tangent line to the function's graph at that point. More simply, it shows us how the function is changing at that point.
When a function is depicted as a curve on a graph, the derivative can determine whether the curve is moving upwards, indicating an increase, or downwards, indicating a decrease.

• For example, if the derivative is positive at a certain point, the function is increasing there.
• If the derivative is negative, the function is decreasing.
• If the derivative equals zero, you might be at a peak, valley, or a point of inflection.

In the example given, the function is expressed as:\[ f(x) = 5 - 4x - x^2 \]Taking the derivative gives us:\[ f'(x) = -4 - 2x \]
This derivative helps identify the intervals where the function is increasing or decreasing.

Critical Points

Critical points in calculus are determined by looking at the function's derivative. They occur where the derivative is zero or undefined, indicating possible peaks or troughs.
These points are important because they help us figure out where the function changes its direction from increasing to decreasing or vice versa.

• To find the critical points, set the derivative equal to zero and solve for \(x\).

In our example, the derivative \( f'(x) = -4 - 2x \) is set to zero:\[ -4 - 2x = 0 \]
Solving this, we find the critical point is:\[ x = -2 \]
This is where we check the function's behavior to see if it switches from increasing to decreasing or the other way around.

Concavity

Concavity tells us about the curvature of the graph of the function. It's determined by the sign of the second derivative.
Much like the first derivative helps with determining slope and whether a function is increasing or decreasing, the second derivative helps determine the function's concavity.

• If the second derivative is positive, the function is concave up, like a smile.
• If negative, it’s concave down, like a frown.

For our function:\[ f(x) = 5 - 4x - x^2 \]
The first derivative was:\[ f'(x) = -4 - 2x \]
The second derivative is:\[ f''(x) = -2 \]
This tells us the function is concave down across the entire real line because \( f''(x) = -2 \) remains constant and negative.

Inflection Points

Inflection points occur where a function changes its concavity from up to down or down to up.
They are found by observing where the second derivative changes sign.

• If the sign of the second derivative switches from positive to negative or negative to positive, an inflection point exists.

In our example:\[ f''(x) = -2 \]
The second derivative is constant and does not change sign. Since it remains consistently negative:
There are no inflection points in this function.
This constant behavior in the second derivative informs us that the shape of the graph is a single concave curve, without any switches in concavity.