Question

Which of the following functions are one-to-one? Select all that apply.

Answer 1

1,2 and 5 is the answers I think

Answer 2

Answer:

$f(x)=x^3-7\,,\,f(x)=\frac{1}{8x-1}$ are one-to-one

Step-by-step explanation:

A function y = f(x) is said to be one-to-one if $f(x_1)=f(x_2)\Rightarrow x_1=x_2$

$f(x)=x^3-7$

$f(x_1)=f(x_2)\\x_1^3-7=x_2^3-7\\x_1^3=x_2^3\\x_1=x_2$

So, f is one-to-one.

$f(x)=x^2-4$

$f(1)=1^2-4=1-4=-3\\f(-1)=(-1)^2-4=1-4=-3\\\Rightarrow f(1)=f(-1)\\\text{but}\,\,1\neq -1$

So, f is not one-to-one

$f(x)=\frac{1}{8x-1}$

$f(x_1)=f(x_2)\\\frac{1}{8x_1-1}=\frac{1}{8x_2-1}\\8x_1-1=8x_2-1\\8x_1=8x_2\\x_1=x_2$

So, f is one-to-one

$f(x)=\frac{5}{x^4}$

$f(1)=\frac{5}{1^4}=5\\f(-1)=\frac{5}{(-1)^4}=5\\f(1)=f(-1)\,\,but\\,\,1\neq -1$

So, f is not one-to-one

$f(x)=\left | x \right |$

$f(1)=\left | 1 \right |=1\\f(-1)=\left | -1 \right |=1\\f(1)=f(-1)\,\,but\,\,1\neq -1$

So, f is not one-to-one

Therefore, functions $f(x)=x^3-7\,,\,f(x)=\frac{1}{8x-1}$ are one-to-one