# Given f(x) = (x^2 + 4mx - 1)/(x + 1), what is m if the critical value of f(x) is x =1.

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### 2 Answers

f(x) = (x^2 + 4mx - 1)/(x + 1), to detrmine m if the critical value of f(x) is x =1.

To find the critical values of f(x).

f(x) = (x^2+4mx-1)/(x+1).

f'(x) = {(x^2+4mx-1)' (x+1)- (x^2+4mx-1)(x+1)'}/(x+1)^2.

f'(x) ={ (2x+4m)(x+1) - (x^2+4mx-1)}/(x+1)^2.

f'(x) = 0 gives: (2x+4m)(x+1) - x^2-4mx+1 = 0.

2x^2 +4mx +2x+4m - x^2-4mx+1 = 0.

x^2 +2x+1+4m = 0.

(x+1)^2 +4m = 0 .

If x= 1 is a critical point, then (1+1)^2+4m = 0. Or 4m = -4.

Therefore m = -4/4 = -1.

Therefore x= 1 is a critical point if m = -1.

The critical value of a function is the root of the first derivative of the function.

We'll determine the first derivative of f(x).

f'(x) = [(x^2 + 4mx - 1)/(x + 1)]'

We'll aply the quotient rule:

f'(x)=[(x^2 + 4mx - 1)'*(x + 1) - (x^2 + 4mx - 1)*(x + 1)']/(x + 1)^2

f'(x) = [(2x+4m)(x+1)-x^2-4mx+1]/(x + 1)^2

We'll remove the brackets:

f'(x) = (2x^2 + 2x + 4mx + 4m - x^2-4mx+1)/(x + 1)^2

We'll combine and eliminate like terms:

f'(x) = (x^2 + 2x +4m + 1)/(x + 1)^2

For x = 1 => f'(1) = 0

f'(1) = (1^2 + 2 +4m + 1)/(1 + 1)^2

(4m+4)/4 = 0

We'll factorize by 4 the numerator:

4(m+1)/4 = 0

We'll simplify and we'll get:

m+1 = 0

**m = -1**

**For m = -1, the function has the critical value x =1.**