Question

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(x)<0f^{\prime}(x)<0 is either strictly positive or strictly negative throughout (a, b). (b) Use the result in (a) to show that x34x+1=0x^{3}-4 x+1=0 has exactly one solution in [−1, 1].

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(a)

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

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