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
$\displaystyle f^{\prime \prime}(x) > 0 $ on the interval I, then f is concave up on I
$\displaystyle f^{\prime \prime}(x) < 0 $ on the interval I, then f is concave down on I
$\displaystyle f(x)= 10x^3 - 3x^5$
$\displaystyle f^\prime = 30x^2 - 15x^4 $
$\displaystyle f^{\prime \prime} = 60x - 60x^3$
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
$\displaystyle f^{\prime \prime} = 0$
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!
Yup,
so let us check
$\displaystyle f^{\prime \prime} = 60x - 60x^3 = 0$
This of course becomes
$\displaystyle 1 - x^2 = 0$
which has the roots
$\displaystyle x=1$ and $\displaystyle x=-1$
EDIT-- I didn't factor correctly. What we have here is
$\displaystyle f^{\prime \prime} = 60x - 60x^3 = 0 = x(1-x^2)$
This introduces the root $\displaystyle x=0$
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
$\displaystyle x < -1$ is part 1
$\displaystyle -1 < x < 0$ is part 2
$\displaystyle 0 < x < 1 $ is part 3
$\displaystyle 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
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
$\displaystyle
f^{\prime \prime} = 60x - 60x^3 = 0
$
Is the same as saying
$\displaystyle f^{\prime \prime} = x(1-x^2) = 0 $
This equation is also satisfied for $\displaystyle x=0$
Therefore our segments become
$\displaystyle x< -1 $ Part 1
$\displaystyle -1 < x < 0$ Part 2
$\displaystyle 0 < x < 1$ Part 3
$\displaystyle x>1$ 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