Verify conditions of Rolle' s theorem
Given: f(x)=x2+2x−8,x∈[−4,2]
Rolle's theorem is satisfied if
(i) f(x) is continuous in [a,b]; f(x)=x2+2x−8 is a polynomial of degree 'two'. so, f(x) is continuous in [−4,2]
(ii) f(x) is differentiable in (a,b); f(x)=x2+2x−8 is a polynomial of degree 'two'. so,f(x) is differentiable in (−4,2)
(iii) f(a)=f(b),f(x)=x2+2x−8
f(−4)=(−4)2+2(−4)−8
⇒f(−4)=16−8−8=16−16=0
f(2)=(2)2+2(2)−8
⇒f(2)=4+4−8=8−8=0
Here, f(−4)=f(2)
Hence, Function is satisfying the conditions of Rolle's theorem.
Finding ′c′
Let ′c′ be the any point in the interval [−4,2], such that f′(c)=0
f(x)=x2+2x−8
f′(x)=2x+2−0
Putting x=c,f′(c)=2c+2
Since all 3 conditions are satisfied so f′(c)=0
2c+2=0
2c=−2
c=−1
Hence c=−1 ∈(−4,2)
Thus, Rolle's theorem is verified.