+0

# Domain and Range

-1
106
3
+193

What is the domain of the function $$f(t) = \frac{1}{(t-1)^2+(t+1)^2}~?$$

Feb 4, 2021

#1
0

The function f(t) is defined for all t except 1 and -1, so the domain is (-inf,-1) U (-1,1) U (1,inf).

Feb 4, 2021
#3
0

This is not correct. The entire denominator must yield 0 for a restriction in the domain to exist, not just one portion of it.

Guest Feb 4, 2021
#2
0

Before trying to pinpoint the domain of this function, it is best to simplify completely. The function can be simplified more by expanding and then combining like terms. Let's do that.

$$f(t) = \frac{1}{(t-1)^2+(t+1)^2}\\ f(t) = \frac{1}{t^2 - 2t + 1 + t^2 + 2t + 1}\\ f(t) = \frac{1}{2t^2 + 2}$$

This expression, while looking slightly different, is the same function algebraically speaking. When considering the domain, it is best to think creatively. What values could cause $$f(t)$$ to be undefined? For rational functions, the only potential for a restriction of the domain is when the denominator yields 0.

$$2t^2+2\neq 0\\ 2t^2 \neq -2$$

I decided not to solve anymore because there every value for t that will satisfy this equation. $$2t^2$$ is always nonnegative, regardless of the input, so it is impossible for $$2t^2$$ to output $$-2$$. Therefore, every value for $$t$$ will yield a real output.

Represented in interval notation,$$Dom(t): (-\infty,+\infty)$$

Feb 4, 2021