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Question

Given an interval $$[a,b]$$ that satisfies hypothesis of Rolle's theorem for the function $$f(x)=x^{4}+x^{2}-2$$. It is known that $$a=-1$$. Find the value of $$b$$.


Solution

$$f(x)=x^{4}+x^{2}-2$$
Since the hypothesis of Rolle's theorem are satisfied by $$f$$ in the interval $$[a,b]$$, we have
$$f(a)=f(b)$$, where $$a=-1$$
Now, $$f(a)$$
$$=f(-1)$$
$$=(-1)4+(-1)2-2$$
$$=1+1-2$$
$$=0$$
and $$f(b)$$
$$=b^{4}+b^2-2$$
$$\therefore f(a)=f(b)$$ gives
$$0=b^{4}+b^{2}-2$$
i.e. $$b^{4}+b^{2}-2=0$$
Since, $$b=1$$ satisfies this equation, $$b=1$$ is one of the root of this equation.
Hence, $$b=1$$.

Mathematics

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