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## Stewart's Calculus 8th Ed.: Question 2, Page 19 Answer Explanations

Question: If \$f(x) = (x^2-x)/(x-1)\$ and \$g(x)=x\$, is it true that f=g?

Page in 8th Edition: 19

Short Answer: No, \$f≠g\$. For input \$x = 1\$, \$f(1)\$ is undefined since the denominator is zero, whereas \$g(1) = 1\$.

Motivated Answer: We’re asked if \$f = g\$, but the equations for \$f\$ and \$g\$ look very different. You might be tempted to say, “No the functions are different because the equations look different.” However, functions can be the same even if the equations look very different.

Remember, functions take in inputs, and spit out outputs. Two functions are only equal if they always give you the same output for a given input.

You can’t manually test every possible input value using pen and paper, since it would take up too much time! But you can try to test whether the equations are the same. So we can write:

\$\$(x^2-x)/(x-1) =? x\$\$

(Here =? means, we’re not sure yet if it’s equal or not.)

Now if you’ve taken algebra before, you might recognize that you can write this as

\$\$(x^2-x)/(x-1)=(x(x-1))/(x-1) =? x\$\$

It’s tempting to cancel out the \$(x-1)\$ and conclude that \$x(x-1)/(x-1) = x\$, but this is not perfectly true.

Remember when you cancel things out from the top and bottom of the fraction, the bottom cannot equal zero. This means that we have the caveat here is that \$x-1≠ 0\$.

If \$x – 1≠ 0\$ in this cancellation, then \$x≠1\$. This gives us the clue we need to get the answer, which is that we can try to put 1 into both equations.

\$f(1)=(1^2– 1)/(1-1)\$ , so \$f\$ is not defined at \$x=1\$.

\$g(1) = 1\$, so \$g\$ is defined at \$x=1\$.

Now, we’ve proven that \$f≠g\$. Remember, if we can find for any input \$x\$, the functions \$f\$ and \$g\$ give different outputs, then \$f\$ and \$g\$ are different, no matter how similar \$f\$ and \$g\$ are for other inputs!

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