Question

f(x)=sinx+cosx-1 x ∈[0,π/2],Verify Rolle's theorem

Answer 1

Answer:

We have given

f(x) = sinx + cosx - 1, x ∈ [0, π/2]

Now differentiate f(x) w. r. t. x

f'(x) = cosx - sinx

f(x) is differentiable on open interval(0, π/2) and continuous on closed interval[0, π/2]

Now, f(a) = f(0) = sin0 + cos0 - 1 =0 + 1 - 1 = 0

f(b) = f(π/2) = sinπ/2 + cosπ/2 - 1= 1 + 0 - 1 = 0

f(a) = f(b) = 0

Thus, function satisfy all the condition of Rolle's Theorem.

Now we have to show that there exist some c∈(0, π/2) such that f'(c) = 0

f(x) = sinx + cosx - 1

Now differentiate f(x) w. r. t. x

f'(x) = cosx - sinx

f'(c) = cosc - sinc =0

cosc = sinc

c = π/4 ∈[0, π/2]

Hence, Rolle's theorem is verified.