# What is Lagrange's applied to function f(x)=2x-1/x+5, -1<=x<=1? interval [-1;1]

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### 1 Answer

You should know that Lagrange's theorem is applied to continuous and differentiable functions.

Since the function is continuous over [-1,1] and it is differentiable over (-1,1), hence, there is a value `c in (-1,1)` such that:

`f(1) - f(-1) = f'(c)(1-(-1))`

`(2-1)/(1+5) - (2-(-1))/(-1+5) = f'(c)(1+1)`

`1/6 - 3/4 = 2f'(c)`

You need to use quotient rule to find f'(x) such that:

`f'(x) = ((2x-1)'(x+5) - (2x-1)(x+5)')/((x+5)^2)`

`f'(x) = ((2(x+5) -2x + 1))/((x+5)^2)`

`f'(x) = (2x + 10 - 2x + 1)/((x+5)^2)`

`f'(x) = 11/((x+5)^2) => f'(c) = 11/((c+5)^2)`

`1/6 - 3/4 = 2f'(c) => (2 - 9)/12 = 22/((c+5)^2)`

`-7/12 = 22/((c+5)^2) => 12*22 = -7((c+5)^2)`

`264 = -7c^2 - 70c - 175`

`7c^2+ 70c + 264 + 175 = 0`

`7c^2 + 70c + 439 = 0`

You need to use quadratic formula such that:

`c_(1,2) = (-70+-sqrt(4900 - 12292))/14`

**Since `sqrt(-7392)!in R` , hence, there is no value for `c in (-1,1)` , hence, the Lagrange's theorem cannot be applied to the given function.**