Thread: Querry: Relations and functions

1. Querry: Relations and functions

Hello people , I was recently studying relations and functions and encountered some questions that i wasn't able to solve. Please help me solve the following questions.

1) Find the range of the functions:

(a) f(x) = 4(x^2) - 4x + 3, x E R
(b) f(x) + 4(x^2) - 4x + 8, x E R

2) If f(x) = (x + 3 / 4x - 5) and y= 3 + 5x / 4x -1 , show that f(y) = x

3) If f(x) = (x - 1 / x + 1); x is not equal to -1, show that f( x -1 / x + 1) = -1 / x

4) If f(x) = x^2 + 3x -2, -1 ≤ x ≤ 1, find the value of x for which f(x) = f(2x + 1)

5) If f(x) = x / 2 and g(x) = 2x + 1, show that 2g [f(x)] = 2f[g(x)] + 1

2. Originally Posted by saberteeth
Hello people , I was recently studying relations and functions and encountered some questions that i wasn't able to solve. Please help me solve the following questions.

1) Find the range of the functions:

(a) f(x) = 4(x^2) - 4x + 3, x E R
(b) f(x) + 4(x^2) - 4x + 8, x E R

2) If f(x) = (x + 3 / x - 5) and y= 3 + 5x / 4x -1 , show that f(y) = x

3) If f(x) = (x - 1 / x + 1); x is not equal to -1, show that f( x -1 / x + 1) = -1 / x

4) If f(x) = x^2 + 3x -2, -1 ≤ x ≤ 1, find the value of x for which f(x) = f(2x + 1)

5) If f(x) = x / 2 and g(x) = 2x + 1, show that 2g [f(x)] = 2f[g(x)] + 1
1) a) $\displaystyle f(x) = 4x^2 - 4x + 3$

$\displaystyle = 4\left(x^2 - x + \frac{3}{4}\right)$

$\displaystyle = 4\left[x^2 - x + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 + \frac{3}{4}\right]$

$\displaystyle = 4\left[\left(x - \frac{1}{2}\right)^2 + \frac{1}{2}\right]$

$\displaystyle = 4\left(x - \frac{1}{2}\right)^2 + 2$.

Notice that $\displaystyle a = 4$. Since this is positive, the turning point is a minimum.

The turning point is $\displaystyle (x, y) = \left(\frac{1}{2}, 2\right)$.

Therefore the minimum value of y is 2.

Thus the range is $\displaystyle y \geq 2$ or $\displaystyle y \in [2, \infty)$.

Do the same process for 1 b).

3. 3) Replace x with $\displaystyle \frac{x-1}{x+1}$:

$\displaystyle f\left(\frac{x-1}{x+1}\right)=\frac{\displaystyle\frac{x-1}{x+1}-1}{\displaystyle\frac{x-1}{x+1}+1}=-\frac{1}{x}$

4) $\displaystyle x^2+3x-2=(2x+1)^2+3(2x+1)-2$

Expand and solve the quadratic.

5) $\displaystyle g(f(x))=g\left(\frac{x}{2}\right)=2\cdot\frac{x}{2 }+1=x+1$

$\displaystyle f(g(x))=f(2x+1)=\frac{2x+1}{2}$

Now prove the equality.

4. Cool!

5. I made an error while typing the 2nd problem in the above list which is now corrected. can you answer that too?