Problem 223 - Calculus Volume 1

Question

For the following exercises, determine a. intervals where \(f\) is concave up or concave down, and b. the inflection points of \(f .\) $$ f(x)=x^{3}-4 x^{2}+x+2 $$

Step-by-Step Solution

Answer

The function is concave down on \((-\infty, \frac{4}{3})\) and concave up on \((\frac{4}{3}, \infty)\), with an inflection point at \((\frac{4}{3}, \frac{8}{27})\).

Step 1: Find the First Derivative

To determine concavity, we first need the first derivative of the function. The first derivative of \( f(x) = x^3 - 4x^2 + x + 2 \) is found using the power rule: \( f'(x) = 3x^2 - 8x + 1 \).

Step 2: Find the Second Derivative

The concavity of the function is determined by its second derivative. Take the derivative of \( f'(x) = 3x^2 - 8x + 1 \).This gives \( f''(x) = 6x - 8 \).

Step 3: Determine Critical Points of the Second Derivative

Set the second derivative equal to zero to find potential inflection points: \( 6x - 8 = 0 \).Solve for \( x \) to find \( x = \frac{4}{3} \).

Step 4: Test Concavity Using Second Derivative Test

Determine intervals using \( x = \frac{4}{3} \) and test the concavity in each interval:- Test for \( x < \frac{4}{3} \): Choose \( x = 0 \), \( f''(0) = 6(0) - 8 = -8 \). Thus, \( f \) is concave down on \(( -\infty , \frac{4}{3})\).- Test for \( x > \frac{4}{3} \): Choose \( x = 2 \), \( f''(2) = 6(2) - 8 = 4 \). Thus, \( f \) is concave up on \(( \frac{4}{3}, \infty )\).

Step 5: Determine Inflection Points

An inflection point occurs where the second derivative changes sign. Since the sign of \( f''(x) \) changes at \( x = \frac{4}{3} \), there is an inflection point at \( x = \frac{4}{3} \).Substitute \( x = \frac{4}{3} \) into \( f(x) \) to find the y-value: \( f(\frac{4}{3}) = (\frac{4}{3})^3 - 4(\frac{4}{3})^2 + \frac{4}{3} + 2 = \frac{8}{27} \).Thus, the inflection point is \( (\frac{4}{3}, \frac{8}{27}) \).

Key Concepts

First Derivative

To understand the concepts of concavity and inflection points, we start with the first derivative of a function. The first derivative, denoted as \( f'(x) \), is essentially the slope or the rate of change of the original function \( f(x) \). In simple terms, it tells us how quickly the function is rising or falling at any given point.

The first derivative is crucial when analyzing a function because it helps identify critical points where the function might have local maxima or minima. In our exercise, we found the first derivative of \( f(x) = x^3 - 4x^2 + x + 2 \) to be \( f'(x) = 3x^2 - 8x + 1 \).

To take this derivative, we apply the power rule where the derivative of \( x^n \) is \( nx^{n-1} \). Understanding this process allows us to find where the slope of the function changes, leading us to insights about its concavity.

Second Derivative

Once the first derivative is found, the next step is to take the second derivative, \( f''(x) \). The second derivative provides insights into the concavity of the function, telling us whether the graph of the function is curving upwards or downwards.

In our exercise, we derived \( f''(x) = 6x - 8 \) from the first derivative. Looking at the value of the second derivative can easily demonstrate concavity:

• If \( f''(x) > 0 \), the function is concave up, resembling a U-shape where the slope is increasing.
• If \( f''(x) < 0 \), the function is concave down, resembling an upside-down U where the slope is decreasing.

By assessing these intervals separately, we found that the function is concave down on \(( -\infty , \frac{4}{3})\) and concave up on \(( \frac{4}{3}, \infty )\). This kind of analysis helps in predicting the behavior of the function's graph more intuitively.

Inflection Point

An inflection point occurs where the function changes its concavity, moving from concave up to concave down or vice versa. This is a spot on the graph where the curve switches its bend, and it happens where the second derivative equals zero and changes its sign.

In the example provided, we identified \( x = \frac{4}{3} \) as a potential inflection point by setting \( 6x - 8 = 0 \) and solving for \( x \). To confirm it as an actual inflection point, we checked that the second derivative changes sign around this point.

Finally, by substituting \( x = \frac{4}{3} \) into the original function \( f(x) \), we found the corresponding \( y \)-coordinate, giving us the inflection point \( (\frac{4}{3}, \frac{8}{27}) \). Recognizing inflection points is vital as they indicate where the curve's nature shifts, providing critical insights in calculus and curve sketching.