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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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