# Thread: Concavity

1. ## Concavity

Short problem. Determine where the given function is concave up and where it is concave down.

f(x)= 10x^3 - 3x^5

Thanks.

2. Originally Posted by akuczma86
Short problem. Determine where the given function is concave up and where it is concave down.

f(x)= 10x^3 - 3x^5

Thanks.
Concavity is shown by the second derivative. Find the second derivative and go from there.

If

$f^{\prime \prime}(x) > 0$ on the interval I, then f is concave up on I

$f^{\prime \prime}(x) < 0$ on the interval I, then f is concave down on I

3. first derivative = 30x^2 - 15x^4.
2nd deriv = 60x - 60x^3.

4. Originally Posted by akuczma86
first derivative = 30x^2 - 15x^4.
2nd deriv = 60x - 60x^3.
$f(x)= 10x^3 - 3x^5$

$f^\prime = 30x^2 - 15x^4$

$f^{\prime \prime} = 60x - 60x^3$

Yup that is correct.

So now, if you want to find where it is concaving up and down. Find the points of inflection, i.e. when

$f^{\prime \prime} = 0$

Then simply plug in a point inbetween the points of inflection to see if it's increasing or decreasing. Then you know if it's concaving up or down!

5. I could be wrong on this, but let's say I plug in 1 and -1. 60(1) - 60(1)^3 = 60-60=0. 60(-1) - 60(-1)^3 = -60 - (-60) = 0. Am I on the right track here?

6. Originally Posted by akuczma86
I could be wrong on this, but let's say I plug in 1 and -1. 60(1) - 60(1)^3 = 60-60=0. 60(-1) - 60(-1)^3 = -60 - (-60) = 0. Am I on the right track here?
Yup,

so let us check

$f^{\prime \prime} = 60x - 60x^3 = 0$

This of course becomes

$1 - x^2 = 0$

which has the roots

$x=1$ and $x=-1$

EDIT-- I didn't factor correctly. What we have here is

$f^{\prime \prime} = 60x - 60x^3 = 0 = x(1-x^2)$

This introduces the root $x=0$

So we know at these points the graph has an inflextion point. What you need to do now is sub in a point inbetween these bounds and outside of them.

In other words, there are 4 parts of this graph. We seperate them into the bounds like

$x < -1$ is part 1

$-1 < x < 0$ is part 2

$0 < x < 1$ is part 3

$x > 1$ is part 4

Sub in a value of x that falls within each of these parts into your second derivative. This will produce the concavity of all 3 intervals

7. So for part 1, 60(-2) - 60(-2)^3 = 60.
Part 2 i re-plugged in 1 and -1.
Part 3 60(2) - 60(2)^3) = -60.

So function is concave up hen x < -1 and the function is concave down when x > 1?

8. Originally Posted by akuczma86
So for part 1, 60(-2) - 60(-2)^3 = 60.
Part 2 i re-plugged in 1 and -1.
Part 3 60(2) - 60(2)^3) = -60.

So function is concave up hen x < -1 and the function is concave down when x > 1?
You can't plug in 1 and -1 for part 2. We already know the function is 0 here! We want to know how the function behaves inbetween these bounds.

Also, I believe I messed up, we have 4 segments not 3! If you notice

$
f^{\prime \prime} = 60x - 60x^3 = 0
$

Is the same as saying

$f^{\prime \prime} = x(1-x^2) = 0$

This equation is also satisfied for $x=0$

Therefore our segments become

$x< -1$ Part 1
$-1 < x < 0$ Part 2
$0 < x < 1$ Part 3
$x>1$ Part 4

My bad!

But the same thing applies here. We must plug numbers inbetween these points. So for example, plug in -100 for part 1, -.5 for part 2, .5 for part 3 and 100 for part 4