In my book, it is given that a function is one-to-one if its derivative is strictly monotonic. Otherwise, it is many-to-one. If this is the case, if we consider a function say f(x)= 2x + sin(x), the derivative is f '(x)= 2 + cos(x). The derivative is clearly non-monotonic, so the given function must be many-to-one. But when I used Desmos to plot the graph, I found it to be one-to-one and strictly increasing. How this is so. Please explain. Is the statement in my book incorrect?
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5$\begingroup$ The function is strictly increasing since the derivative is always positive. So if you're quoting your book accurately, then the book is wrong. Which book is it? $\endgroup$– quasiCommented Aug 2, 2019 at 8:25
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3$\begingroup$ I suggest that you read again what your textbook says and, it if it really says that, then get another textbook. If $f(x)=x^2$, then $f$ is not one-to-one, but $f'(x)=2x$, which is strictly monotonic. $\endgroup$– José Carlos SantosCommented Aug 2, 2019 at 8:27
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1$\begingroup$ A function is one-to-one if each value in the range is given by ONLY ONE value in the domain. So strictly monotone FUNCTIONS are one-to-one. Since strictly monotone is implied by having a strictly positive/negative DERIVATIVE, these functions are monotone also. Now, since $|\cos x| \leq 1$, your derivative there is strictly positive, so your function is strictly increasing and therefore one-to-one. $\endgroup$– DzoooksCommented Aug 2, 2019 at 8:28
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1$\begingroup$ If the function itself is strictly monotonic, then it is one-to-one. A differentiable function is strictly monotonic if its derivative is strictly positive (or possibly $0$ at isolated points). $\endgroup$– ArthurCommented Aug 2, 2019 at 8:32
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1$\begingroup$ @Dzoooks Thank you for your explanation $\endgroup$– VishnuCommented Aug 2, 2019 at 8:38
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1 Answer
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Let's stick to continuous, differentiable functions. A real function $f$ is one-to-one if it is strictly monotonic (the function itself, not its derivaitve!!)
You can guarantee that a function is strictly monotonic if its derivative does not change sign (let's ignore zeros for now. I mean: it either stays positive, or stays negative). In your case, $f'(x)=2+\cos{(x)}$ is always greater than zero.
It could be the case that $f'$ has a zero and $f$ is still strictly monotonic (think of $f(x)=x^3$), but what cannot happen is that there are points where $f'$ is positive and other points where $f'$ is negative
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$\begingroup$ Thank you for your answer David. If a function is one-to-one then its slope must be either positive or negative. The slope of the function may be zero at selective points only. Then we might say that the derivative of the function is monotonic right? Please explain if I have misunderstood. Sorry for disturbing you again. $\endgroup$– VishnuCommented Aug 2, 2019 at 8:53
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1$\begingroup$ No. There are plenty of functions that always have the same sign but are not monotonic (think again about the derivaitve of $x^3$ as an example) For instance, the derivative of your original function is always positive, but not monotonic at all! $\endgroup$– DavidCommented Aug 2, 2019 at 8:55
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1$\begingroup$ Thanks a lot David. So finally, if we want the derivative to be monotonic then the second derivative of the function must be positive. Further, monotonicity of the derivative of a function does not affect whether a function is one-to-one or many to one. Just the sign of the derivative determines the fact. Please tell whether I am correct now? $\endgroup$– VishnuCommented Aug 2, 2019 at 9:01
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1$\begingroup$ @GuruVishnu That's right! It could also be the case that the second derivaitve is negative. Then the first derivative would be strictly decreasing $\endgroup$– DavidCommented Aug 2, 2019 at 9:02
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