Differentiation question

Does anybody know how to find stationary point and maximum and minimum for a curve using differentiation? If you do can you please explain it to me thanks

Differentiating once will give you the gradient function. At a turning point, the gradient of the curve is 0, so you set this gradient function to 0 and solve for x to find the x co-ord then sub into the original equation to find the y co-ord. You do not really need to determine the max or min from differentiation as you can do it from thinking about how the graph looks like and sketching it. However if you want to, you simply differentiate the curve again and plug in the x co-ord of the stationary point. This new function will tell you the rate of change of gradient. If you get a negative value then it means the rate of change of gradient is negative both sides of that point meaning that it is a maximum point. If you get a positive value it means the rate of change of gradient is positive both sides of that point meaning it is a minimum point. If you get 0 it means it is a point of inflection I believe.

Hey thanks for all that detail. Can you maybe show me an example of this for the equation 2x^2+5x-6?

Sure, ill write it on a piece of paper and post it, give me a moment.

I'll show you it for $y=x^2+2x+1$. This is $y=(x+1)^2$, and so we already know that the minimum occurs at $x=-1$, but I'll show you how to check it using differentiation.

We know that at a maximum or minimum, the gradient must be 0, as otherwise a point to the left or right could be larger/ smaller. So we want to differentiate this and set it's derivative equal to 0.

$\frac{dy}{dx} = 2x+2 = 0 \implies x=-1$.

Substituting this into y, we get $y(-1) = 0$, which is what we expected to get.

I'll show you it for $y=x^2+2x+1$. (The spoiler contains extra information that isn't necessary for this question).We know that at a maximum or minimum, the gradient must be 0, as otherwise a point to the left or right could be larger/ smaller. So we want to differentiate this and set it's derivative equal to 0.

$\frac{dy}{dx} = 2x+2 = 0 \implies x=-1$.

Substituting this into y, we get $y(-1) = 0$.

Why are you confusing him with completed square form and vertices? Clearly needs to grasp the basics before progressing onto that.


Posted from TSR Mobile

Thanks a lot man

Have you tried using exam solutions.net, the guy is extremely good at explaining things. Also what level are you learning this at?


Posted from TSR Mobile

I hope this helps!

I put effort into trying to explain it along the way as best as I can.

Posted from TSR Mobile