Calculus How to find the points of the graph!

Calculus How to find the points of the graph!

I have here a function:

f(x)=(x+1)^(2)(x-5)^(3)(x+4)^(4)(x-3)^(5)

I need help finding all the turning points of that function.

I need help finding all the inflection points.

Can you guys please include working out, so i can understand it.

Thanks!

Re: Calculus How to find the points of the graph!

Hey Jeka.

Turning points have a derivative equal to zero and inflexion points have an extra condition where the second derivative is zero.

So you need to find out where the first derivative is zero and then test whether the second derivative at that point is also zero. If it's not zero (i.e the second derivative) then you have a normal turning point.

To understand this intuitively, the second derivative tells how the first derivative is changing: if its positive it means the slope is increasing and if its negative it means the slope is decreasing. At a turning point something "turns around" so it means the slope will continue to "decrease" or "increase" but if it's zero it means that it's not going to continue to "turn" as it has been up to the turning point and if this is the case, it's a point of inflexion (another name is called a saddle point).

Re: Calculus How to find the points of the graph!

Hey Chiro.

I have a question, when you find the first and second derivative, do i split the function into four parts? as in:

f(x)=(x+1)^(2)(x-5)^(3)(x+4)^(4)(x-3)^(5)

to

f(x)=(x+1)^(2)

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

f(x)=(x+4)^(4)

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

Do i do that?

Or do i find the first and second derivative as the whole function?

And if i change actually change it into parts

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

i get the first derivative as

f'(x)=5x^4-60x^3+270x^2-540x+405

and how do i find 'x'?

And, how do you find the turning points? Because in this function there are 2 types of graph, 2 parabola's and 2 cubic graphs

How do i find where they intercept?

And thats it for now! Sorry for so much questions

Re: Calculus How to find the points of the graph!

You need to find it for the whole function which should be a 14 degree polynomial.

You can either expand it out or you can use the product rule where d(uv)/dx = u*(dv/dx) + v*(du/dx)

Re: Calculus How to find the points of the graph!

What do you do after that?