Question

LMVT says that if y = f(x) be a given function which is ;

a.Continuous in [a,b]

b. Differentiable in (a,b)

Then, f'(c) = $\frac{f\left(b\right)-f\left(a\right)}{b-a}$ for some c $ϵ$ (a,b)

Find the value of c for the function f(x) =$-x^2$+4x-5 and the interval [-1,1]

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Answer 1

Since the given function is polynomial, we can say that the two conditions of LMVT, functions is continuous and differentiable, are satisfied.

To find c, we will use the relation f'(c) = $\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Here, f'(x) = -2x+4

$\Rightarrow$ f'(c)=-2c+4

f(a) = f(-1) = -10

f(b) = f(1) = -2

b-a = 1-(-1) = 2

f'(c) = $\frac{f\left(b\right)-f\left(a\right)}{b-a}\Rightarrow$ -2c+4 = $\frac{-2-(-10)}{2}$ = 4

$\Rightarrow c=0$