Problem 24

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. $$f(x)=x^{2} \ln x$$

Step-by-Step Solution

Verified

Answer

(a) Increasing on $(e^{-1/2}, \infty)$. (b) Decreasing on $(0, e^{-1/2})$. (c) Concave up on $(e^{-3/2}, \infty)$. (d) Concave down on $(0, e^{-3/2})$. (e) Inflection point at $x = e^{-3/2}$.

1Step 1: Find the Derivative of f(x)

Start by finding the first derivative of the function, $ f(x) = x^2 \ln x $. Use the product rule for differentiation, which states that if you have two functions multiplied together, $ u(x) \cdot v(x) $, the derivative is $ u'(x)v(x) + u(x)v'(x) $. Here, $ u(x) = x^2 $ and $ v(x) = \ln x $. Find $ u'(x) $ and $ v'(x) $:\[ u'(x) = 2x, \quad v'(x) = \frac{1}{x} \]Now apply the product rule:\[ f'(x) = (x^2)' \cdot \ln x + x^2 \cdot (\ln x)' = 2x \ln x + x \]

2Step 2: Find Critical Points

To find where $ f(x) $ is increasing or decreasing, find the critical points by setting $ f'(x) = 0 $:\[ 2x \ln x + x = 0 \]\[ x(2 \ln x + 1) = 0 \]This gives us the solutions:1. $ x = 0 $2. $ 2 \ln x + 1 = 0 \Rightarrow \ln x = -\frac{1}{2} \Rightarrow x = e^{-1/2} \approx 0.606 $We exclude $ x = 0 $ in our domain because $ \ln x $ is undefined there.

3Step 3: Test Intervals Around Critical Points for Increasing/Decreasing Behavior

Use the critical points to create test intervals: $(0, e^{-1/2})$ and $(e^{-1/2}, \infty)$. Choose a test point in each interval, such as $ x = 0.5 $ and $ x = 1 $, to determine the sign of $ f'(x) $.- For $ x = 0.5 $, $ f'(0.5) = 2(0.5) \ln(0.5) + 0.5 < 0 $, so $ f(x) $ is decreasing on $(0, e^{-1/2})$.- For $ x = 1 $, $ f'(1) = 2(1) \ln(1) + 1 = 1 > 0 $, so $ f(x) $ is increasing on $(e^{-1/2}, \infty)$.

4Step 4: Find the Second Derivative of f(x)

Differentiate $ f'(x) = 2x \ln x + x $ to find the second derivative:Use the product rule on $ 2x \ln x $:\[ (2x \ln x)' = 2 \ln x + 2 \]Thus,\[ f''(x) = 2 \ln x + 2 + 1 = 2 \ln x + 3 \]

5Step 5: Determine Concavity

Set $ f''(x) = 0 $ to find potential inflection points:\[ 2 \ln x + 3 = 0 \Rightarrow \ln x = -\frac{3}{2} \Rightarrow x = e^{-3/2} \approx 0.223 \]Test intervals around $ x = e^{-3/2} $ for concavity:- For $ x < e^{-3/2} $, choose $ x = 0.1 $. $ f''(0.1) = 2 \ln(0.1) + 3 < 0 $, so the interval is concave down.- For $ x > e^{-3/2} $, choose $ x = 1 $. $ f''(1) = 2 \ln(1) + 3 > 0 $, so the interval is concave up.

6Step 6: Conclusion

Using the information:- $ f(x) $ is increasing on $(e^{-1/2}, \infty)$.- $ f(x) $ is decreasing on $(0, e^{-1/2})$.- $ f(x) $ is concave up on $(e^{-3/2}, \infty)$.- $ f(x) $ is concave down on $(0, e^{-3/2})$.- The inflection point is at $ x = e^{-3/2} $.

Key Concepts

DerivativeConcavityCritical PointsInflection Points

Derivative

In calculus, derivatives represent the rate at which a function changes as its input changes. For the function $f(x) = x^2 \ln x$, finding the derivative involves applying the product rule. It's crucial because it helps us understand how the function behaves, particularly whether it is increasing or decreasing in certain intervals.

The product rule states that the derivative of a product $u(x) \cdot v(x)$ is $u'(x)v(x) + u(x)v'(x)$.
For our function, $u(x) = x^2$ and $v(x) = \ln x$.
Differentiate to get $u'(x) = 2x$ and $v'(x) = \frac{1}{x}$.
Applying the product rule, the derivative $f'(x) = 2x \ln x + x$.

This derivative tells us how the function $f(x)$ behaves and is used to find critical points.

Concavity

Concavity describes the direction in which a curve bends. It depends on the second derivative of a function. We determine concavity by analyzing the second derivative, $f''(x)$.

If $f''(x) > 0$, the function is concave up (like a smiley face).
If $f''(x) < 0$, the function is concave down (like a frown).

For $f(x) = x^2 \ln x$, we first find the second derivative:

Differentiate the first derivative $f'(x) = 2x \ln x + x$ to get $f''(x) = 2 \ln x + 3$.

By examining $f''(x)$, we can identify where the function is concave up or down. This analysis is crucial for understanding the complete behavior of $f(x)$ over its domain.

Critical Points

Critical points are the $x$-values where the derivative is zero or undefined. They signify points where the function's slope changes direction, indicating possible maxima, minima, or points of inflection.

For a function $f(x)$, find critical points by solving $f'(x) = 0$.
Plug in the factorized derivative to get $x(2 \ln x + 1) = 0$, leading to $x = e^{-1/2} \approx 0.606$.

To determine the nature of these points:

Test intervals around critical points to see if the function is increasing or decreasing.
If the function changes from increasing to decreasing, it indicates a local maximum.
If it changes from decreasing to increasing, it signifies a local minimum.

Identifying these critical points helps us know more about the function's graph and its turning behaviors.

Inflection Points

Inflection points occur where the concavity of a function changes. This is different from simply having a critical point. To find inflection points, we set the second derivative equal to zero and solve: