## Related questions with answers

We frequently must solve equations of the form f(x) = 0. When f is a continuous function on [a, b] and f(a) and f(b) have opposite signs, the intermediate-value theorem guarantees that there exists at least one solution of the equation f(x) = 0 in [a, b]. (a) Explain in words why there exists exactly one solution in (a, b) if, in addition, f is differentiable in (a, b) and $f^{\prime}(x)<0$ is either strictly positive or strictly negative throughout (a, b). (b) Use the result in (a) to show that $x^{3}-4 x+1=0$ has exactly one solution in [−1, 1].

Solution

Verified(a)

If $f'(x)>0,\ \forall x \in (a,b) \ \Rightarrow \ f$ is an increasing function on $(a,b) \ \Rightarrow \ f$ is one to one mapping, that means, any $y$-value has exactly one $x$-value such that $f(x)=y$. Here, for $y=0$, there is exactly one $x$ such that $f(x)=0$. The same conclusion is obtained if $f'(x)<0, \ \forall x \in (a,b)$ (except, $f$ is than decreasing).

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