## One-to-one functions: how would finding derivative prove it?

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### One-to-one functions: how would finding derivative prove it?

Consider the function $f(x)\, =\, x^4\, -\, 2x^3\, +\, 1$

1. Show that $f$ is one-to-one on the interval $\left[\frac{3}{2},\, \infty\right)$.
Hint: Show that the derivative $f'$ is positive on $\left[\frac{3}{2},\, \infty\right)$.
...
2. Explain briefly why $f$ is not one-to-one on $\left(-\infty,\, \infty\right).$

Could you please explain this problem to me? I don't understand why finding the derivative f' (ie, the gradient at a point) would prove that it is a one-to-one function.

Thanks
moriarty

Posts: 1
Joined: Thu Mar 26, 2009 7:58 am

For a function to be one-to-one, there will be no y-value that is reached from two or more x-values. For instance, a linear equation such as y = 3x - 5 is one-to-one, because each y-value is reached from only one x-value. On the other hand, y = sin(x) is not one-to-one, because y = 0 for x = 0, pi, 2pi, 3pi, etc, etc. Many x-values go to the one y-value, so the function is many-to-one rather than one-to-one.

If a function's derivative is always positive, then can there be any spot at which the function "dips back down" to repeat an earlier y-value?

You might be expected to use Rolle's Theorem or something...? Supposing f(a) = f(b) for some "a" not equal to "b", then g(x) = f(x) - f(a) would have the property that g(a) = g(b) = 0. Then there exists some "c" between "a" and "b" so g'(c) = 0 = f'(c). But what will you have shown about f'(x) for all x, including this hypothetical "c"?

stapel_eliz

Posts: 1803
Joined: Mon Dec 08, 2008 4:22 pm