State whether it has any points at which it is not differentiable
I have no idea what to do...
Lets propose a hypothesis and test any potential doubts:
Proposal:
Since all points that are sharp or are discontinous are non-differentiable, and our function here is a piece-wise function. That is, its a function composed of many other functions all with given domains.
The proposal is that at x=-2 and x=3, we have either discontinous or sharp points.
We can tell that they might be sharp because at x=2 and x=-3 there is a 'transition' between two functions, that is its a starting point of one of teh functions and we can
predict that they might be sharp.
However we must test whether this is actually true and to do this we must observe the gradients of the functions at these 'transition' points.
Lets look at the point x=3 for instance. Its the point which connects the graph f(x)=2x and f(x)=3
By observation they are both straight lines, moreover they are straight lines with
different gradients.
How can we tell?
For f(x)=2x, due to y=mx+b our m=2, and hence our gradient is 2
For f(x)=3, its a horizontal line hence gradient of 0
So at x=3 we have a connection between the two straight lines, but they are different gradients, different slopes HENCE they MUST have a sharp point connecting them (you cant connect two lines in a striaght line if they are of different slopes obviously)
Hence at x=3 is a sharp point (or it may be discontinuous, the point is no matter how it connects IF it does it cannot connect smoothly), HENCE it is a point
which is non-differentiable
Lets look at x=-2 now, its a bit more interesting since our graph y=1-x^2 has a varying gradient so IT COULD connect smoothly to the striaght line. However lets think about this, we are connecting a straight line graph to a parabola.
So that means that
in order for it to be smooth our straight line must connect with the parabola
at its turning point. Why? because at its turning point the curve is flat (you should of covered this in the logic behind calculus)
So to find the turning point we must differentiate f(x)=1-x^2. But I assume you havent done this yet, so lets take the rote-y approach and rearrange this into our general parabola form:
So our turning point by formula is x=-b/2a, b=0 in this case, hence our turning point is at x=0
But hang on, the point we are originally caring about here and the point we are questioning is x=-2, so that means the turning point is NOT x=-2, that means when the horizontal line intersects the parabola it must do so at a sharp point.
Hence at x=-2 and x=3 the points are non-differentiable since they are sharp/discontinous. We have achieved the question without graphing.
