Problem 62 - Algebra and Trigonometry

Question

62–63 ? Find, correct 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

(a) The function increases on \((-\infty, \frac{1}{2})\) and decreases on \((\frac{1}{2}, \infty)\). (b) Its range is \((0, 10^{1/4}]\).

Step 1: Identify the Derivative

To determine where the function increases or decreases, we need to find the derivative of the function. The function given is \( y = 10^{x - x^2} \). Using the chain rule, the derivative is: \[ y' = 10^{x - x^2} \cdot \ln(10) \cdot (1 - 2x) \] where \( \ln \) denotes the natural logarithm.

Step 2: Find Critical Points

Set the derivative equal to zero to find critical points: \[ 10^{x - x^2} \cdot \ln(10) \cdot (1 - 2x) = 0 \]Since \( 10^{x - x^2} \) is never zero, \( \ln(10) eq 0 \), we solve \( 1 - 2x = 0 \).Thus, \( x = \frac{1}{2} \) is a critical point.

Step 3: Test Intervals for Increasing/Decreasing

Test the intervals around the critical point \( x = \frac{1}{2} \):- For \( x < \frac{1}{2} \), choose \( x = 0 \) and find \( y'(0) = 10^{0} \cdot \ln(10) \cdot 1 = \ln(10) \), which is positive.- For \( x > \frac{1}{2} \), choose \( x = 1 \) and find \( y'(1) = 10^{0} \cdot \ln(10) \cdot (-1) = -\ln(10) \), which is negative.Thus, the function is increasing on \(( -\infty, \frac{1}{2} )\) and decreasing on \(( \frac{1}{2}, \infty )\).

Step 4: Determine the Range

Since the function \( y = 10^{x-x^2} \) has a maximum point at \( x = \frac{1}{2} \), find the maximum value:\[ y\left(\frac{1}{2}\right) = 10^{\frac{1}{2} - (\frac{1}{2})^2} = 10^{\frac{1}{2} - \frac{1}{4}} = 10^{\frac{1}{4}} \]The value is greater than 0 for all \( x \), so the range of the function is \((0, 10^{1/4}]\).

Key Concepts

Derivative

In calculus, the derivative of a function gives us the rate at which the function's value changes concerning changes in the input, or the variable. For the given function, \( y = 10^{x-x^2} \), finding the derivative involves applying the chain rule, a crucial tool for differentiating composite functions. The composite function has two parts: the exponential part \( 10^{(...)} \) and the inner function \( x-x^2 \).
Utilize the chain rule as follows:

• Differentiate the outer function: \( 10^{x-x^2} \) becomes \( 10^{x-x^2} \cdot \ln(10) \)
• Differentiate the inner function: \( x-x^2 \) gives \( 1 - 2x \)

These parts combine to form the derivative of the original function: \[y' = 10^{x-x^2} \cdot \ln(10) \cdot (1-2x) \]With this derivative, you can analyze the function's behavior, such as identifying critical points and determining where it increases or decreases.

Increasing and Decreasing Intervals

A critical aspect of understanding the behavior of functions is identifying intervals where they are increasing or decreasing. To do this, we use the sign of the derivative. If the derivative \( y' \) is positive on an interval, the function is increasing there. Conversely, if \( y' \) is negative, the function is decreasing.
For the function \( y = 10^{x-x^2} \), we have determined its derivative as \( y' = 10^{x-x^2} \cdot \ln(10) \cdot (1-2x) \). To find where the function changes from increasing to decreasing, locate the critical points, then test intervals around these points using the sign of \( y' \).

• To the left of \( x = \frac{1}{2} \), test \( x = 0 \): \( y'(0) > 0 \), function increases.
• To the right of \( x = \frac{1}{2} \), test \( x = 1 \): \( y'(1) < 0 \), function decreases.

Thus, the function increases on the interval \( (-\infty, \frac{1}{2}) \) and decreases on \( (\frac{1}{2}, \infty) \). Recognizing these intervals helps graph the function and understand its trend over different sections of the domain.

Range of a Function

The range of a function is the set of all possible output values (\( y \)-values) generated by the input values of the domain. Determining the range is particularly meaningful for exponential functions like \( y = 10^{x-x^2} \) because their behavior changes based on input, \( x \). To find the range:

• Identify extreme values or limits: Given its increasing and decreasing nature, locate the function's maximum value at its critical point \( x = \frac{1}{2} \).
• Plug \( x = \frac{1}{2} \) into the original function: \( y\left( \frac{1}{2}\right) = 10^{\frac{1}{4}} \)

Thus, at \( x = \frac{1}{2} \), the function achieves its maximum, indicating the upper bound of the range. Given the function's behavior and its exponential base, the lowest value is slightly above zero, and the upper limit is \( 10^{\frac{1}{4}} \). Consequently, the range is \( (0, 10^{1/4}] \). This range reflects the function's output potential over its domain.

Critical Points

Critical points in a function occur where its derivative is zero or undefined, indicating potential local maxima, minima, or points of inflection. For \( y = 10^{x-x^2} \), find critical points by setting its derivative equal to zero: \[10^{x-x^2} \cdot \ln(10) \cdot (1 - 2x) = 0\]Considering \( 10^{x-x^2} \) and \( \ln(10) \) are never zero, we focus on:

• \( 1 - 2x = 0 \)
• Solving gives \( x = \frac{1}{2} \) as the critical point.

Critical points are crucial for understanding where the function's direction shifts, potentially identifying maximum or minimum points in the interval. By checking \( x = \frac{1}{2} \), the derivative changes sign, confirming it as a point of maximum value, where the function shifts from increasing to decreasing.