Melody

Thanks Chris,

If you  draw them one under the other so that the x values line up, it help with the graphing too.

I sort of did that. 

I hope you understand but you will need to practice to cement it into your brain.

Also you need to think about some of the sideways things I have talked about.  (Mainly in #6 post)

If you can work out their relevance it will help you a real lot.

I just took a photo of some of my scribble, it is not to scale.

Hopefully it will help you.

I added the answer on you question on your post because the site was plying up. Hopefully it will behave itself now.

yes LOL

If you think about it this first one is a cubic graph (degree of 3)     (up down and up again)

When you differentiate a cubic you get a quadratic (with a degree of 2)

Quadratics are parabolas   :)

So the graph of the gradients of a cubic will give a parabolic curve.

\(y=x^3 \qquad \text{cubic}\\ y'=3x^2 \qquad \text{parabolic}\\ y''=6x \qquad \text{Straight line}\\ y'''=6 \qquad \text{horizonal line}\\ y''''=0 \\\)

see if you can work out the relevance of all that. :)

These can be quite tricky.

consider the first one.

when x=1/6 the gradient is 0

when x=-3/2 the gradient is 0.

plot those 2 points first.

now when x>1/6 the gradient is negative and it is getting steeper and steeper.

so as x tends to infinity, the gradient tends to -infinity.  

When x is <-2/3 the same thing is happening

so as x tends to -infinity, the gradient tends to -infinity also.

Now between -2/3 and 1/6 the curve is increasing.  I mean the gradient of the tangent is positive

Somewhere in the middle between the 2 is a point of inflection.  This is where the first derivative is still positive but it is becoming smaller so the graphe between -2/3 and 1/6 will look a bit like a concave down parabola. the point at the top will be the point of inflection of the original function graph.

Do you get all that?  Try drawing it.  I think these are fun to work out.

Then have a go at the second one.

Partial mathematician:

I know you are keen to learn and I really do appreciate that, I want you to stick around for a long time ....  but please stop pretending that you know a whole lot more than you really do.  You leave people confused and sometimes annoyed. Some of your input is really good, I know this, but ....  you can finish this sentence for yourself.

You are very welcome. I will remember your good manners :)

\(f(x)=4\sqrt{x}+\sqrt[4]{x}\\ f(x)=4*x^{0.5}+x^{0.25}\\ f'(x)=2x^{-0.5}+0.25x^{-0.75}\\ f'(16)=2*16^{-0.5}+0.25*16^{-0.75}\\ f'(16)=2*4^{-1}+0.25*16^{3/4*-1}\\ f'(16)=2*4^{-1}+0.25*2^{3*-1}\\ f'(16)=2*4^{-1}+0.25*8^{-1}\\ f'(16)=\frac{2}{4}+\frac{0.25}{8}\\ f'(16)=\frac{16}{32}+\frac{1}{32}\\ f'(16)=\frac{17}{32}\\ \)

The gradient for part b is NOT 5/2

However, it is positive so the curve is increasing.

I really did not need to do all that. As soon as the curve was differentiated it was clear that the for all positive values of x the curve would be increasing.   I mean the first derivative would be positive.