Question

# Verify Rolle's theorem for the function $$f(x)={ x }^{ 2 }+x-6$$ in the interval $$[-3,2]$$.

Solution

## $$f(x)=x^{2}+x-6, x \in [-3,2]$$Since, $$f(x)=x^{2}+x-6$$ is a polynomial and every polynomial function is continuous for all $$x \in R$$So, $$f(x)$$ is continuous at $$x \in [-3,2]$$Also, every polynomial function is differentiable.So, $$f(x)$$ is differentiable at $$(-3,2)$$Now,$$f(-3)=9-3-6=0$$$$f(2)=4+2-6=0$$Hence, $$f(-3)=f(2)=0$$Now, $$f'(x)=2x+1$$$$f'(c)=2c+1$$Since, all the conditions are satisfied,$$f'(c)=0$$$$2c+1=0$$$$c=-\dfrac{1}{2}$$And, $$-\dfrac{1}{2} \in [-3,2]$$Hence, Rolle's theorem is verified.Maths

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