In order to determine the value of of the function we set The value becomes

As such

In order to determine the value of of the function we set The value becomes

As such

The denomiator factors out to:

The denominator becomes zero when . But the function can exist at any other value.

The denomiator factors out to:

The denominator becomes zero when . But the function can exist at any other value.

Simplify:

Even though the cancels out from the numerator and denominator, there is still a hole where the function discontinues at . The function also does not exist at , where the denominator becomes .

Evaluate the following function when

To evaluate this function, simply plug-in 6 for t and simplify:

So our answer is:

Simplify:

Even though the cancels out from the numerator and denominator, there is still a hole where the function discontinues at . The function also does not exist at , where the denominator becomes .

Evaluate the following function when

To evaluate this function, simply plug-in 6 for t and simplify:

So our answer is:

The square cannot house any negative term or can the denominator be zero. So the lower limit is since cannot be , but any value greater than it is ok. And the upper limit is infinity.

Find the value of the following function when

To evaluate this function, plug in 2 for x everywhere it arises and simplify:

So our answer must be undefined, because we cannot divide by