Problem 45 - Precalculus Mathematics for Calculus

Question

Find, rounded to two decimal places, (a) the intervals on which the function is increasing or decreasing and (b) the range of the function. $$y=10^{x-x^{2}}$$

Step-by-Step Solution

Answer

Function increases on \((-\infty, \frac{1}{2})\), decreases on \((\frac{1}{2}, \infty)\); range is \((0, 1.78]\).

Step 1: Find the derivative

To determine where the function is increasing or decreasing, we first find the derivative of the function. The original function is given by \( y = 10^{x - x^2} \). By the chain rule, the derivative \( y' \) is \( y' = 10^{x-x^2} \cdot \ln(10) \cdot (1 - 2x) \).

Step 2: Locate critical points

Set the derivative \( y' \) equal to zero to find the critical points: \( 10^{x-x^2} \cdot \ln(10) \cdot (1 - 2x) = 0 \). This simplifies to \( 1 - 2x = 0 \), so \( x = \frac{1}{2} \).

Step 3: Determine intervals of increase and decrease

To determine the increasing or decreasing nature of \( y \), evaluate the sign of the derivative in the intervals \((-\infty, \frac{1}{2})\) and \((\frac{1}{2}, \infty)\). For \(x < \frac{1}{2}\), \(1 - 2x > 0\), so \(y' > 0\) (function is increasing). For \(x > \frac{1}{2}\), \(1 - 2x < 0\), so \(y' < 0\) (function is decreasing).

Step 4: Define intervals of increase and decrease

The function \( y = 10^{x-x^2} \) is increasing on the interval \((-\infty, \frac{1}{2})\) and decreasing on the interval \((\frac{1}{2}, \infty)\).

Step 5: Find the range of the function

Analyze the behavior of the function as \(x\) approaches different critical points. As \(x \to -\infty\), \(10^{x-x^2} \to 0\). As \(x \to \frac{1}{2}\), the function reaches a peak because this point gives the maximum value when the function changes from increasing to decreasing. The value of the function at \(x = \frac{1}{2}\) is \(10^{\frac{1}{2} - (\frac{1}{2})^2} = 10^{\frac{1}{4}}\). As \(x \to \infty\), \(10^{x-x^2} \to 0\). Thus, the range of the function is \((0, 10^{1/4}]\).

Step 6: Calculate numerical values

To confirm the precision, calculate \(10^{1/4}\) using a calculator to find a numerical approximation. \(10^{1/4} \approx 1.77828\), so the range of the function is approximately \((0, 1.78]\).

Key Concepts

Derivative

Derivatives are a fundamental concept in calculus. They measure how a function changes as its input changes. For the function \( y = 10^{x-x^2} \), the derivative helps us find where the function is increasing or decreasing.
To find the derivative of this function, we apply the chain rule. The chain rule is a method for finding the rate of change of compositions of functions. For \( y = 10^{x-x^2} \), the outer function is \( 10^u \) and the inner function \( u = x-x^2 \). Differentiating it gives:

• The derivative of the outer function \( 10^u \) is \( 10^u \ln(10) \).
• The derivative of the inner function \( x-x^2 \) is \( 1 - 2x \).

Combining these, the derivative \( y' = 10^{x-x^2} \cdot \ln(10) \cdot (1 - 2x) \).
This expression is crucial for determining how the function behaves at different intervals.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points are valuable because they often indicate where the function's behavior changes, such as switching from increasing to decreasing.
For the function \( y = 10^{x-x^2} \), we found its derivative to be \( y' = 10^{x-x^2} \cdot \ln(10) \cdot (1 - 2x) \).
To find critical points, we set the derivative equal to zero: \[10^{x-x^2} \cdot \ln(10) \cdot (1 - 2x) = 0\]Since \( 10^{x-x^2} \cdot \ln(10) \) is never zero, \( 1 - 2x = 0 \) gives the critical point \( x = \frac{1}{2} \).
At \( x = \frac{1}{2} \), the nature of the function changes, making it an important location for analyzing the function.

Intervals of Increase and Decrease

Knowing the intervals where a function increases or decreases helps in graphing and understanding its behavior. We use the sign of the derivative to determine these intervals.
For \( y = 10^{x-x^2} \), its derivative is positive or negative depending on the interval around the critical point \( x = \frac{1}{2} \). Considering:

• For \( x < \frac{1}{2} \), \( 1 - 2x > 0 \) means \( y' > 0 \), indicating the function is increasing.
• For \( x > \frac{1}{2} \), \( 1 - 2x < 0 \) means \( y' < 0 \), showing the function is decreasing.

Thus, we conclude that \( y = 10^{x-x^2} \) is increasing over \((-\infty, \frac{1}{2})\) and decreasing over \((\frac{1}{2}, \infty)\).
These intervals describe how the function moves over the x-axis, either climbing up or down.

Range of a Function

The range of a function describes the set of possible output values (y-values). It provides insight into how the function behaves vertically.
For the function \( y = 10^{x-x^2} \), understanding the range involves analyzing the behavior of the function as \( x \) approaches its limits and the critical point.

• As \( x \to -\infty \), the expression \( x-x^2 \) becomes very negative, making \( 10^{x-x^2} \to 0 \).
• As \( x \to \infty \), \( x-x^2 \) becomes negative again, so \( 10^{x-x^2} \to 0 \).
• At \( x = \frac{1}{2} \), the function reaches its maximum value of \( 10^{\frac{1}{4}} \), approximately 1.78.

Therefore, the range of the function is \((0, 10^{1/4}]\), or approximately \((0, 1.78]\), meaning the function outputs values from just above 0 up to approximately 1.78.