Equal functions

Problem : Determine whether f(x) and g(x) are identical functions?

$$f\left(x\right)=x$$ $$g\left(x\right)=\sqrt{x^2}$$

Solution : Here, f(x) is defined for all values of x and its domain is R. On the other hand, domain of g(x) is also R as square of x is always non-negative. However, square root of a number is non-negative. Therefore, two function forms are not equivalent as f(x) is real, whereas is g(x) is non-negative and is a subset of R. Thus, range of f(x) is R and range of g(x) is (0,∞). Clearly, two given functions are not equal.

Problem : Determine whether f(x) and g(x) are identical functions.

$$f\left(x\right)=\frac{x}{x^2}$$ $$g\left(x\right)=\frac{1}{x}$$

Solution : Two function forms are equivalent as f(x) is reduced to g(x) on simplification. Now, expression of f(x) is defined for all values of x except x=0. Thus, domain of f(x) is R-{0}. On the other hand, domain of reciprocal function g(x) is also R-{0}. Clearly, two given functions are equal.

Problem : 3. Determine whether f(x) and g(x) are identical functions.

$$f\left(x\right)={\log }_{e}x-{\log }_e\left(x^2+1\right)$$ $$g\left(x\right)=\log {}_e\left(\frac{x}{1+x^2}\right)$$

Solution :

Two function forms are equivalent as f(x) is changed to g(x) and vice-versa on simplification. Now, f(x) is defined for

$$x>0\text{and}{x}^2+1>0$$

But $x^2$ is always positive. Hence, domain of f(x) is (0, ∞). On the other hand, g(x) is defined for :

$$\frac{x}{1+x^2}>0$$ $$x>0$$

Thus, domain of g(x) is also (0, ∞). Hence, two functions are identical.

Problem : Determine domains for which two functions are equal.

$$f\left(x\right)=\log x-\log \left(x-1\right)$$ $$g\left(x\right)=\log \left(\frac{x}{x-1}\right)$$

Solution : Two function forms are equivalent as f(x) is changed to g(x) and vice-versa on simplification. Now, f(x) is defined for

$$x>0\text{and}x-1>0$$ $$x>0\text{and}x>1$$

Hence, domain of f(x) is intersection of two intervals (1, ∞). On the other hand, g(x) is defined for :

$$\frac{x}{x-1}>0$$

Critical points are 0 and 1. Using sign rule for rational function, the domain of g(x) is values of x satisfying above inequality :

$$\left(-\infty ,0\right)\cup \left(\mathrm{1,}\infty \right)$$

Clearly, two domains are not equal. Note that there is no restriction on the range of the functions. Therefore, two functions are equal in the restricted domain which is intersection of two domains.

$$\text{Domain}=\left(\mathrm{1,}\infty \right)$$