Another extrapolation

Algebra Level 4

Let $f,g: \mathbb{R} \longrightarrow \mathbb{R}$ be two continuous functions.

What do the graphs $\frac{f(x)}{g(x)}$ and $f(x)-g(x)$ say about one another?

A: If $f(x)-g(x) = 0$ then $\frac{f(x)}{g(x)} = 1$ .

B: If $\frac{f(x)}{g(x)}$ is monotonically increasing on some interval, then $f(x)-g(x)$ is positive on that interval.

C: None of the above.

A only A & B B only C

1 solution

Jake Lai
May 23, 2015

A: It may be that $f(x) = g(x) = 0$ and so $\frac{f(x)}{g(x)}$ is undefined.

B: A quick counterexample is when $f(x) = g(x)+x = 2x+1$ .

Hence, only C is true.

For (A), you can also take $f(x)=g(x)$ as some function such that $g(x^*)=0$ for some $x^*\in\Bbb R$ . There are infinitely many functions that does the job (for example, any polynomial function with real roots).

Take, for example, $f(x)=g(x)=(x-1)(x-2)(x-3)$ . This can act as a counter-example for both the claims. For (A), $f(x)-g(x)=0$ but $\frac{f(x)}{g(x)}$ is undefined at all the roots, i.e., at $x=1,2,3$ . Now, for (B), take any interval that doesn't contain the roots. $\frac{f(x)}{g(x)}=1$ for those intervals and is hence monotonically increasing (constant function here). And we have $f(x)-g(x)=0\not\in\Bbb{R^+}$

Prasun Biswas - 6 years ago

Nice! Good observations.

Jake Lai - 6 years ago

Another extrapolation

1 solution

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