Fraternal functions

Algebra Level 2

Suppose f ( x ) = g ( x ) 4 x 5 f(x) = \frac{g(x)}{4x-5} and f ( x ) = g ( x ) 3 x 2 f'(x) = \frac{g(x)}{3x-2} . Find the value of m m that satisfies f f ( m ) = 5 ff'(m) = -5 given that g ( x ) = x + 4 g(x) = x+4 .

This is a modified high school problem.


The answer is 3.

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1 solution

Noel Lo
May 10, 2015

Method 1:

f f ( m ) = 5 ff'(m) = -5

f ( g ( m ) 3 m 2 ) = 5 f(\frac{g(m)}{3m-2}) = -5

f ( m + 4 3 m 2 ) = 5 f(\frac{m+4}{3m-2}) = -5

g ( m + 4 3 m 2 ) 4 ( m + 4 3 m 2 ) 5 = 5 \frac{g(\frac{m+4}{3m-2})}{4(\frac{m+4}{3m-2})-5} = -5

m + 4 3 m 2 + 4 4 ( m + 4 3 m 2 ) 5 = 5 \frac{\frac{m+4}{3m-2}+4}{4(\frac{m+4}{3m-2}) -5} = -5

m + 4 + 12 m 8 4 m + 16 ( 15 m 10 ) = 5 \frac{m+4+12m-8}{4m+16-(15m-10)}=-5

13 m 4 26 11 m = 5 \frac{13m-4}{26-11m} = -5

13 m 4 = 55 m 130 13m-4 = 55m - 130

( 55 13 ) m = 130 4 (55-13)m = 130-4

42 m = 126 42m = 126

m = 126 42 = 3 m = \frac{126}{42}=\boxed{3}

Method 2:

f f ( m ) = 5 ff'(m) = -5

g f ( m ) 4 f ( m ) 5 = 5 \frac{gf'(m)}{4f'(m)-5} = -5

f ( m ) + 4 4 f ( m ) 5 = 5 \frac{f'(m) + 4}{4f'(m) - 5} = -5

f ( m ) + 4 = 25 20 f ( m ) f'(m)+4 = 25-20f'(m)

21 f ( m ) = 21 21f'(m)=21

f ( m ) = 1 f'(m)=1

g ( m ) 3 m 2 = 1 \frac{g(m)}{3m-2} = 1

m + 4 3 m 2 = 1 \frac{m+4}{3m-2} = 1

m + 4 = 3 m 2 m+4 = 3m-2

2 m = 6 2m = 6

m = 3 m=\boxed{3}

Hi, I don't think this should be under calculus. I believe it should be under algebra.

Noel Lo - 6 years, 1 month ago

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