Maths Help ? ?

Question

a) f(x) = 1 - 1/x find f(f(1/2))

b) if 2^3x+2/4^2x+1 = 8^x-2 find the value of x

c) Find all the possible values of x that satisfy x + 2 + sqrt x+2 = 56

Answer 1

a)

f(x) = 1 - 1/x

f(1/2) = 1 - 1 / (1/2) = 1 - 2 = -1

f(f(1/2)) = f(-2) = 1 - 1/-1 = 1 + 1 = 2

***************

b)

2^(3x + 2) / 4^(2x + 1) = 8^(x - 2)

Put everything into base 2.

4^(2x + 1) = (2^2)^(2x + 1) = 2^2(2x + 1) = 2^(4x + 2)

8^(x - 2) = (2^3)^(x - 2) = 2^(3x - 6)

Now you have...

2^(3x + 2) / 2^(4x + 2) = 2^(3x - 6)

2^(3x + 2 - (4x + 2)) = 2^(3x - 6)

2^(3x + 2 - 4x - 2) = 2^(3x - 5)

2^(-x) = 2^(3x - 5)

Now you can drop the 2s and just work on the exponents...

-x = 3x - 5

-4x = -5

x = 5/4

***********

c)

x + 2 + sqrt (x + 2) = 56

sqrt (x + 2) = 54 - x

square both sides....

x + 2 = (54 - x)^2

x + 2 = 2916 - 108x + x^2

0 = x^2 - 109x + 2914

Use quadratic formula to finish solving for x.

Answer 2

ok...this is a little revision for me.

a) f(f(x))= 1-1/(1-1/x)

f(f(1/2))= 1-1/(1-1/(1/2))

          = 1-1/-1

          =1-1 = 0

answer is 0

ill do next one in a mo

Answer 3

a) f(1/2) = 1 - 2 = -1

therefore f(-1) = 1 + 1 = 2,

therefore f(f(1/2)) = 2

b) add some parentheses I cant quite tell where the x is placed, especially on the 2nd term

c) x + 2 + sqrt(x+2) = 56

let sqrt(x + 2) = y

y^2 + y = 56,

y^2 + y - 56 = 0, solve

(y + 8)(y - 7) = 0

y = -8, y = 7

sqrt(x + 2) = -8, does not exist, a sqrt cannot be negative;

therefore  sqrt(x + 2) = 7, solve

x + 2 = 49

x = 47