questions involving domain and range...
1. Find the domain of the function
f(x)=(3x+1)/(sqrt of (x^2+x-2))
andd.
2. Let f(x)= (absvalue(x))/x.
Show that f(x)={ 1, x > 0
{-1, x < 0
Find the domain and range of f(x).
To find the domain of a function, you need to determine the values of x for which the function is defined.
1. Find the domain of the function f(x) = (3x + 1) / √(x^2 + x - 2):
The domain is the set of all possible x-values that the function can take. In this case, there are two things to consider:
a) The denominator, √(x^2 + x - 2), must not equal zero because division by zero is undefined. So, we need to find the values of x that make the denominator zero.
√(x^2 + x - 2) = 0
(x^2 + x - 2) = 0
Solving this quadratic equation, we get x = -2 and x = 1 as the values that make the denominator zero. Therefore, x cannot be equal to -2 or 1.
b) Apart from the values that make the denominator zero, the function is defined for all other values of x.
Therefore, the domain of f(x) is all real numbers except x = -2 and x = 1.
2. Let f(x) = |x| / x.
To show that f(x) = 1 for x > 0 and f(x) = -1 for x < 0, we need to consider the properties of the absolute value function.
The absolute value of any positive number is itself, so |x| = x for x > 0.
The absolute value of any negative number is the negation of itself, so |x| = -x for x < 0.
Hence, for x > 0, f(x) = |x| / x = x / x = 1.
And for x < 0, f(x) = |x| / x = -x / x = -1.
To find the domain and range of f(x):
The domain is the set of all possible x-values for which the function is defined. In this case, the denominator x cannot be zero because division by zero is undefined. Therefore, the domain of f(x) is all real numbers except x = 0.
The range is the set of all possible y-values that the function can take. In this case, since f(x) can only be 1 or -1, the range of f(x) is {1, -1}.
So, the domain is all real numbers except x = 0, and the range is {1, -1}.