91Ó°ÊÓ

Derivatives are a fundamental concept in calculus, used to determine the rate at which a function is changing at any given point. When applied to a polynomial function like \(y = 10^{x-x^2}\), finding the derivative helps us understand how the function behaves over different intervals.
In this case, we calculate the derivative using the chain rule, a technique used when differentiating composite functions. Our initial function can be rewritten as \(y = 10^u\) by letting \(u = x - x^2\). To find the derivative \(\frac{dy}{dx}\), apply the chain rule: differentiate the outer function first, \(10^u \ln(10)\), and multiply it by the derivative of the inner function, \(1-2x\). Consequently, \(\frac{dy}{dx} = 10^{x - x^2} \ln(10) (1-2x)\).
This derivative gives us a tool to analyze whether the function is increasing or decreasing at any point based on the sign of \(\frac{dy}{dx}\). A positive derivative indicates an increasing function, while a negative derivative indicates a decreasing one.

To identify the intervals where a function is increasing or decreasing, we examine the sign of its derivative. For the function \(y = 10^{x-x^2}\), the derivative \(\frac{dy}{dx} = 10^{x-x^2} \ln(10) (1-2x)\) determines this behavior.
By setting \(\frac{dy}{dx} > 0\), the function is increasing. This simplification leads us to solve \(\ln(10) (1-2x) > 0\), arriving at the condition \(1-2x > 0\). Solving gives \(x < 0.5\), indicating the function is increasing in the interval \(( -\infty, 0.5 )\).
Conversely, setting \(\frac{dy}{dx} < 0\) helps find where the function is decreasing, yielding \(1-2x < 0\). Solving for \(x\) leads to \(x > 0.5\), so the function decreases on \((0.5, \infty)\). Recognizing these intervals helps understand how the function behaves across its domain.

The range of a function includes all possible output values, or \(y\) values, that the function can produce. For the function \(y = 10^{x-x^2}\), determining the range involves analyzing the expression \(x - x^2\), which describes a parabola opening downward.
The vertex of this parabola provides the maximum value of \(x-x^2\), located at \(x = 0.5\). Calculating at the vertex gives \(0.5 - (0.5)^2 = 0.25\). Therefore, the maximum output for \(y\) is \(10^{0.25}\), achieved at \(x = 0.5\).
As \(x\) approaches \(\pm \infty\), the term \(x-x^2\) tends to \(-\infty\), driving \(10^{x-x^2}\) towards zero. Consequently, the function range is \((0, 10^{0.25}]\), indicating \(y\) values span from just above zero up to \(10^{0.25}\). Understanding the range offers insights into the potential outputs of the function depending on \(x\) values.