Question

Find the value of c if all the condition of Rolle's theorem satisfies for function
f(x)=x2+x-6 in (-3, 2]
0
-2
-1/2
-3/2​

Answer 1

c=0

Rolle's theorem states that if a feature f is non-stop at the closed interval [a, b] and differentiable at the open interval (a, b) such that f(a) = f(b), then f′(x) = zero for a few x with a ≤ x ≤ b.

Rolle’s Theorem is a selected case of the suggested polynomial theorem which satisfies positive conditions. At the identical time, Lagrange’s suggest price theorem is the suggest price theorem itself or the primary suggest price theorem.  In general, you possibly can recognize suggest because the common of the given values. But withinside the case of integrals, the method of locating the suggest price of  specific capabilities is specific. Let us study Rolle’s theorem and the suggest price of such capabilities and their geometrical interpretation.

Lagrange’s Mean Value Theorem

If a feature f  is described at the closed interval [a,b] pleasurable the subsequent conditions –

i) The feature f is non-stop at the closed interval [a, b]

ii)The feature f  is differentiable at the open interval (a, b)

Then there exists a price  x =  c in the sort of manner that

f'(c) = [f(b) – f(a)]/(b-a)

This theorem is likewise called the primary suggest price theorem or Lagrange’s suggest price theorem.

f(x)=$x^{2} +x^{-6}$

being a polynomial function is everywhere continuous and differentiable. Also,

f(−1)=f(1)=0.

⇒

           From Rolle’s theorem, there exists at least one c such that f '(c) = 0.

Again, we see that there are two such c’s is given by  f′(c)=0

$2c^{2} -(-6c^{2} )=0$

$2c^{2} +6c^{2} =0\\8c^{2} =0\\c=\frac{0}{8} \\c=0$