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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More