直线斜率取值范围

Analyzing function behavior

I'm looking at the function $y = \frac{x+1}{|x|+1}$ to see if a line passes through three distinct points with their x-coordinates summing to zero. This leads me to explore the slope possibilities.

Mapping the function

Breaking down $y = \frac{x+1}{|x|+1}$ into two cases: for $x \ge 0$, $y = 1$; and for $x < 0$, $y = \frac{x+1}{1-x}$.

Charting the curve

For $x < 0$, I’m analyzing $y = \frac{x+1}{1-x}$. Calculating specific values and observing limits to understand the function's behavior better.

Piecing together intersections

Considering a line $y = kx + b$ that intersects $y = 1$ for $x \geq 0$ and $y = \frac{x+1}{1-x}$ for $x < 0$.

Finding intersections

I’m hypothesizing that to achieve three distinct intersections, the line slopes in a way that it intersects $y = \frac{x+1}{1-x}$ at two points for $x < 0$ and $y = 1$ for $x \geq 0$.

Solving quadratic equations

I'm working through the quadratic equation $m x^2 + x(1 - m + c) + (1 - c) = 0$ to ensure distinct negative solutions for $x < 0$.

Ensuring intersection validity

I’m confirming the discriminant is positive and both x-values for $x < 0$ are negative. Checking the sum of x-coordinates to ensure the balance of intersections is maintained.

Balancing equations

Ensuring the sum of the x-coordinates equals zero, I deduced that $c = \frac{m}{2}$ by combining and simplifying terms from the quadratic equation.

Ensuring distinct intersections

I’m connecting $c = \frac{m}{2}$ to maintain the sum of x-coordinates as zero. Now, I’m examining conditions on $m$ to ensure the intersections are distinct and valid.

Laying out the options

OK, I’m working through different cases: for $m > 0$, $0 < m < 2$. $m < 0$ and $m = 0$ lead to contradictions.

Confirming validity

I’m concluding that for the line to intersect in $x > 0$, $0 < m < 2$ holds true. Negative slopes and $m = 0$ lead to contradictions.

Mapping intersections

I’m identifying the intersection points of the line $y = mx + \frac{m}{2}$ with $y = \frac{x+1}{1-x}$, ensuring they are distinct within $x < 0$.

Piecing together the equation

I’m carefully simplifying the expression $(1-x)(x+1/2)$ to ensure all terms are combined correctly, aiming to solve for the intersections within $x < 0$.

Simplifying the equation

I’m combining like terms to streamline the equation $m x^2 + (1 - m/2) x + (1 - m/2)=0$. This involves defining coefficients $A = m$, $B = (1 - m/2)$, and $C = (1 - m/2)$.

Determining intersections

I’m ensuring two distinct real solutions for $x < 0$ by checking the discriminant $p(p - 4m) > 0$. Noting that $p = 1 - \frac{m}{2}$ and $p$ is positive.

Ensuring progress

I'm combining inequalities to confirm $m < \frac{2}{9}$ to guarantee two distinct real solutions for $x < 0$. This maintains positivity and leads to distinct intersections.

Confirming negative solutions

I’m piecing together the roots' sign. With a negative sum and positive product, it’s clear that the two distinct real solutions for $x < 0$ are negative.

Piecing together intersections

I found that for $0 < m < \frac{2}{9}$, the line intersects the function at three distinct points, balancing positive and negative x-coordinates and summing to zero.

Confirming intersection points

To achieve three distinct intersection points, the avenue between $0 < m < \frac{2}{9}$ remains open, showing distinct and valid intersections for both positive and negative $x$-coordinates.

Finding the sweet spot

I’m focusing on $0 < m < \frac{2}{9}$ to ensure three unique intersections with the function, balancing positive and negative $x$-coordinates perfectly.

Analyzing endpoints

I’m thinking about how the slope range $0 < m < \frac{2}{9}$ affects the intersections. For $m = 0$, the line $y = 0$ only intersects at $x = -1$. For $m = \frac{2}{9}$, I’m determining if it intersects the function at three distinct points and if it should be included in the range.

Thought about 函数图像与直线斜率 for 47 seconds