# The intermediate value theorem states the following: Consider an interval I=[a,b] in the real numbers R and a continuous function f:I->R. Then, If u is a number between f(a) and f(b), f(a)<u<f(b) (or f(a)>u>f(b) ), then there is a c∈(a,b) such that f(c)=u.

The intermediate value theorem states the following: Consider an interval $I=\left[a,b\right]$ in the real numbers $ℝ$ and a continuous function $f:I\to ℝ$. Then, If $u$ is a number between $f\left(a\right)$ and $f\left(b\right)$, $f\left(a\right) (or $f\left(a\right)>u>f\left(b\right)$ ),

then there is a $c\in \left(a,b\right)$ such that $f\left(c\right)=u$.
What happens if we consider $\mathbb{R}$ instead of $I$ ?
Is there a simple method to prove the theorem in this case ? I know the method for $f:I\to \mathbb{R}$ with $I=\left[a,b\right]$, but here it is different.
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hitturn35
The general equation for the point-slope form of a linear equation is $y-{y}_{1}=m\left(x-{x}_{1}\right)$, where m is the slope, and $\left({x}_{1},{y}_{1}\right)$ is the known point.
$y-\left(-5\right)=-5\left(x-\left(-3\right)\right)$ =
$y+5=-5\left(x+3\right)$