If a continuous function $$f$$ defined on the real line $$R$$, assumes positive and negative values in $$R$$ then the equation $$f(x)=0$$ has a root in $$R$$. For example, if it is known that a continuous function $$f$$ on $$R$$ is positive at some point and its minimum value is negative then the equation $$f(x)=0$$ has a root in $$R$$.
Consider $$f\left( x \right) = k{e^x} - x$$ for all real $$x$$ where $$k$$ is real constant.

The line $$y=x$$ meets $$y = k{e^x}$$ for $$k \le 0$$ at

If a continuous function $$f$$ defined on the real line $$R$$, assumes positive and negative values in $$R$$ then the equation $$f(x)=0$$ has a root in $$R$$. For example, if it is known that a continuous function $$f$$ on $$R$$ is positive at some point and its minimum value is negative then the equation $$f(x)=0$$ has a root in $$R$$.
Consider $$f\left( x \right) = k{e^x} - x$$ for all real $$x$$ where $$k$$ is real constant.

For $$k>0$$, the set of all values of $$k$$ for which $$k{e^x} - x = 0$$ has two distinct roots is

If a continuous function $$f$$ defined on the real line $$R$$, assumes positive and negative values in $$R$$ then the equation $$f(x)=0$$ has a root in $$R$$. For example, if it is known that a continuous function $$f$$ on $$R$$ is positive at some point and its minimum value is negative then the equation $$f(x)=0$$ has a root in $$R$$.
Consider $$f\left( x \right) = k{e^x} - x$$ for all real $$x$$ where $$k$$ is real constant.

The positive value of $$k$$ for which $$k{e^x} - x = 0$$ has only one root is

If $$P(x)$$ is a polynomial of degree less than or equal to $$2$$ and $$S$$ is the set of all such polynomials so that $$P(0)=0$$, $$P(1)=1$$ and $$P'\left( x \right) > 0\,\,\forall x \in \left[ {0,1} \right],$$ then