Hỏi hàm số y=13x3−x2+x𝑦=13𝑥3−𝑥2+𝑥 đồng biến trên khoảng nào ? A. (−∞;+∞)(−∞;+∞). B. (−∞;1)(−∞;1) và (1;+∞)(1;+∞). C. (−∞;1)(−∞;1). D. (1;+∞)(1;+∞).

1. Break Down the Problem

We need to determine the intervals on which the function $y = 13x^3 - x^2 + x$ is increasing. This involves finding the derivative of the function and analyzing its sign.

2. Relevant Concepts

To find where the function is increasing, we need to compute its first derivative $y'$ and analyze where $y' > 0$.

3. Analysis and Detail

Step 1: Compute the derivative

The function is given as: $y = 13x^3 - x^2 + x$

The derivative of $y$ with respect to $x$ is: $y' = \frac{d}{dx}(13x^3 - x^2 + x) = 39x^2 - 2x + 1$

Step 2: Determine the critical points

To find the critical points, we set the derivative equal to zero: $39x^2 - 2x + 1 = 0$

Applying the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: Where $a = 39$, $b = -2$, and $c = 1$: $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 39 \cdot 1}}{2 \cdot 39}$ $x = \frac{2 \pm \sqrt{4 - 156}}{78}$ $x = \frac{2 \pm \sqrt{-152}}{78}$

Since the discriminant is negative, there are no real roots. Therefore, the derivative $y'$ does not change sign and must be evaluated to determine if it's positive or negative.

Step 3: Analyze the sign of the derivative

Evaluate $y'$:

For $x = 0$: $y'(0) = 39(0)^2 - 2(0) + 1 = 1 > 0$

Since $y' > 0$ throughout the entire real line, the function $y = 13x^3 - x^2 + x$ is increasing everywhere.

4. Verify and Summarize

We verified the calculations, and since there are no critical points where $y' = 0$ or $y' < 0$, we conclude the function is strictly increasing over all real numbers.

Final Answer

The function is increasing on the interval $(-\infty, +\infty)$. Thus, the answer is A. (−∞; +∞).