# Thread: Surjuctivity of a function

1. ## Surjuctivity of a function

Why is the following function surjective(onto) ?

$f: (1,\infty)\longrightarrow (\ln 2,\infty)$
$f(x)=\int_{x^2}^{x^4}\! \frac{1}{\ln t}\, dt$

2. Originally Posted by bigli
Why is the following function surjective(onto) ?

$f: (1,\infty)\longrightarrow (\ln 2,\infty)$
$f(x)=\int_{x^2}^{x^4}\! \frac{1}{\ln t}\, dt$
Try using the IVT and creating some nice inequalities

3. is this IVT? : intermediate value theorem. anyway, How do I use it?

4. Originally Posted by bigli
What is IVT?
Intermediate value theorem.

5. How? please hint me.

6. The interemdiate value theorem says that if f(x) is continous on the interval [a, b] then it takes on all values between f(a) and f(b) on that interval.

Here, f(x) is $\int_{x^2}^{x^4}\frac{1}{ln t} dt$ which is continous on [tex][1, \infty) because ln t is continuous for all positive t and the integral of a continuous function is continuous.

You will still need to show that f(x) is unbounded. That is, that if y is any positive number there exist x such that f(x)> y. That, I think, is why Drexel28 mentions "some nice inequalities".

7. In fact, how do I show $\lim_{x\rightarrow 1^+ }\int_{x^2}^{x^4}\!\frac{1}{\ln t}\, dt=\ln 2$ ?