# Thread: Graphing using calculus

1. ## Graphing using calculus

Hey all, could someone please check my answers? thanks

1.) Consider the function k(x) = (x^2+1)arctan(x)

Useful Information:
k'(x) = 1+2xarctan(x)
k''(x) = (2x/x^2+1)+ 2arctan(x)

__________________________________________________ ______________
a) Find the domain of k

my solution: dom(k) = R

__________________________________________________ ______________

b) Find x and y intercepts

my solution x int at (0,0) and y int at (0,0)

__________________________________________________ ______________

c) Find the interval on which k is increasing:

my solution: since k'(x) > 0 on (0,inf), k increases from (0,inf)

__________________________________________________ ______________

d) Find the interval on which k is decreasing

my solution: since k'(x) < 0 on (-inf,0), k decreases from (-inf, 0)

__________________________________________________ ______________

e) Does k have any stationary points?

my solution: k'(x) = 0
1+2xarctan(x) = 0
arctan(x) = -(1/2x)
since the two graphs never intersect, there are no stationary points

__________________________________________________ _______________

f) show that k has point of inflection at x = 0

my solution: k''(0) = 0, therefore point of inflection.

__________________________________________________ ____________________

g) Is the point of inflection at x = 0 stationary or non stationary?

my solution: k'(0) = 1, therefore non stationary point of inflection at x = 0

__________________________________________________ _____________________

h) Does k have any other point of inflections?

my solution: k''(x) = 0
(2x/1+x^2)+2 arctan(x) = 0
I got stuck at this point, any guidance would be much appreciated.

Thanks.

2. ## Re: Graphing using calculus

This looks fine! There's only one point of inplection $(0,0)$. It's not easy to solve equations like $\frac{2x}{1+x^2}+2\arctan(x)=0$ because you're dealing with polynomial functions and cyclometric functions at the same time. Offcourse by inspection you notice $x=0$ is a zero and this is the only zero and thus point of inflection (take a look at the graph).

3. ## Re: Graphing using calculus

parts (c) & (d) appear to be wrong.

4. ## Re: Graphing using calculus

Originally Posted by SammyS
parts (c) & (d) appear to be wrong.
Without a maximum or a minimum you can't use a sign table to determine where the function decreases or increases. So I used the graph of the first derivative. Why is it not correct?

5. ## Re: Graphing using calculus

Hello Guys.
What I did for part c and d was solve 1 +2xarctan(x) = 0

which i found to be arctan(x) = -1/2x

since arctan is > -1/2x on (0, inf) and < -1/2x on (-inf,0) i figured that would be the points when k was increasing and decreasing respectively

6. ## Re: Graphing using calculus

Originally Posted by andrew2322
Hello Guys.
What I did for part c and d was solve 1 +2xarctan(x) = 0

which i found to be arctan(x) = -1/2x

since arctan is > -1/2x on (0, inf) and < -1/2x on (-inf,0) i figured that would be the points when k was increasing and decreasing respectively
$1 + 2x\arctan{x} \ge 1$ for all $x$ ...

7. ## Re: Graphing using calculus

Originally Posted by skeeter
$1 + 2x\arctan{x} \ge 1$ for all $x$ ...
I dont understand?

8. ## Re: Graphing using calculus

$\arctan{x} =0$ only at x = 0.

So that $2x\arctan{x}=0$ only at x = 0, and $2x\arctan{x}$ is an even function with an absolute minimum of 0 at x = 0.

Therefore, $2x\arctan{x}\ge 0\,.$

This gives you $1+2x\arctan{x}\ge 1\,.$

The first derivative is never zero.