# Convex and concave function

These kind of functions are new to me or at least I’ve not recalled doing these previously.

I need some clarification..

Example question: x³+3x²-2

So I differentiate it and equate to zero and find that the stationary points are 0 and -2…

I understand that convex has the shape of a valley, if I imagine the line of two points above as a water level. So this is above the function line of the valley is my function.

So, I apply the second derivative and find that x=0 tells me that’s a local minimum and x=-2 is a local max….

The material that I am using tells me that a convex interval for all x is when the second derivative >0…..

So, the interval should be x>=0?

I am a little confused because the worked material on here tells me it’s x>=-1 but when I tested that, the answer was 0..

Edit: x=0.5 is positive, ahhh, wait I see my error x>-1 not >=1 🙂

Okay, so x>-1 is correct. It's that simple then?

Their working:

My working:

I drew part of the graph to get a better grasp
(edited 1 month ago)
Original post by KingRich
These kind of functions are new to me or at least I’ve not recalled doing these previously.
I need some clarification..
Example question: x³+3x²-2
So I differentiate it and equate to zero and find that the stationary points are 0 and -2…
I understand that convex has the shape of a valley, if I imagine the line of two points above as a water level. So this is above the function line of the valley is my function.
So, I apply the second derivative and find that x=0 tells me that’s a local minimum and x=-2 is a local max….
The material that I am using tells me that a convex interval for all x is when the second derivative >0…..
So, the interval should be x>0?
I am a little confused because the worked material on here tells me it’s x>=-1 but when I tested that, the answer was 0..

If 2nd deriv > 0 then 6x + 6 > 0 so ....
Original post by Muttley79
If 2nd deriv > 0 then 6x + 6 > 0 so ....

I just edited as I was misreading it
Original post by Muttley79
If 2nd deriv > 0 then 6x + 6 > 0 so ....

Hold a second. I don’t need to find the x routes, do I? As I’ve found the second derivative to be 6x+6, This is this the information required to clarify the interval? As you pointed out 6x+6>0, x>-1,

So if it asked me to find concave it would be x<-1 ?
Original post by KingRich
Hold a second. I don’t need to find the x routes, do I? As I’ve found the second derivative to be 6x+6, This is this the information required to clarify the interval? As you pointed out 6x+6>0, x>-1,
So if it asked me to find concave it would be x<-1 ?

You know a lot from the equation of the curve as the x^3 term is positive, you know the y axis intercept and the stationary points so can sketch from that.

Yes to the question.
Original post by Muttley79
You know a lot from the equation of the curve as the x^3 term is positive, you know the y axis intercept and the stationary points so can sketch from that.
Yes to the question.

For this question attached. As you pointed out, an equation to a graph can give a lot of information.

So, based on that.

Would this be correct?

Original post by KingRich
For this question attached. As you pointed out, an equation to a graph can give a lot of information.
So, based on that.
Would this be correct?

I'd probably write a bit more to show where I got the inequlity values from but looks OK otherwise.
Original post by Muttley79
I'd probably write a bit more to show where I got the inequlity values from but looks OK otherwise.

Nice. I was writing it at the bottom and then I figured huh, I think the equation is self explanatory lol.

I would have applied the factor theorem to find the x
Values for the first derivative.
And then plugged them into the second derivative as I did previously. Is this the best way to approach this question?

I considered using second derivative like the previous question but I wasn’t sure how I’d approach it?

If I solved the x’s and said x=-1 and x=1/3

Could I then just write when x=-1, dy²/dx²<0… etc
Original post by KingRich
Nice. I was writing it at the bottom and then I figured huh, I think the equation is self explanatory lol.
I would have applied the factor theorem to find the x
Values for the first derivative.
And then plugged them into the second derivative as I did previously. Is this the best way to approach this question?
I considered using second derivative like the previous question but I wasn’t sure how I’d approach it?
If I solved the x’s and said x=-1 and x=1/3
Could I then just write when x=-1, dy²/dx²<0… etc

I did it the same wsy as before - you get a quadratic for the second derivative so solve it and get values - then I sketched the quadratic and found when > 0 and when < 0
Original post by Muttley79
I did it the same wsy as before - you get a quadratic for the second derivative so solve it and get values - then I sketched the quadratic and found when > 0 and when < 0

Nice 🙂 thank you for clarification.
Original post by KingRich
Nice 🙂 thank you for clarification.

No problem