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Current time:0:00Total duration:3:35

consider F which is defined for all real numbers and what arguments X is f of X not the differential not differentiable so to think about that I'm actually going to try to visualize what f prime of x must look like so I'm going to do F prime of X in this purple this purple color so if we look at f of X right over here it looks like it's slope its slope is pretty much consistently negative 2 over this interval between x equals I guess it's like negative 8 and 1/2 all the way up to x equals negative 2 it looks like the slope is a constant negative 2 so if I were to draw its derivative its derivative would look something its derivative would look something like this this derivative looks something like this but then something interesting happens at x equals negative 2 right as we cross x equals negative 2 it looks like the slope goes from being negative 2 it looks like the slope goes from being negative to being positive and it looks like right out the get-go if I were to estimate the slope of this tangent line it starts changing it's not a line anymore it's a curve the slope of the tangent line right at this point looks like it's around I don't know it looks like it's around 3 and a half because if I were to draw if I were to draw a tangent line right over here it looks like it looks like if I move one in the X direction I move up about 3 and 1/2 in the Y direction so I'm just trying to obviously estimate it so it looks like the slope goes up to 3 and a half right when I cross that point right when I cross that point and then the slope becomes lower and lower and lower all the way until I get to this point right over here all the way until I get to x equals 2 and it looks like it continues to get lower all the way until you get to x equals 3 so it looks like the slope of the line is and it looks like it's getting lower at a constant rate I guess I could say so it looks like it's doing something like this something like this something like this over this interval over this interval but then right as X crosses 3 this becomes a flat line the slope is 0 here so right as X crosses 3 the slope becomes zero so we immediately see their points where it looks like the slope jumps it looks like the slope jumps and at these points we really don't have a defined derivative the slope jumps there as well and so the this at what arguments is f not differentiable what's not differentiable when X is equal to negative two when X is equal to negative two we really don't have a slope there if we try to find remember when we're trying to find the slope of the tangent line we take the limit of the slope of the secant line between that point and some other point on the curve if we did that as we approach from the left it looks like the derivative is negative two if we do that from the right it looks like the derivative is something like positive three and a half and so we're not getting the same limit of the secant line as as we approach from the left and as we approach from the right and the same thing is happening at X is equal to three at x equals three as we approach from the left the slope looks like it is decreasing and is approaching it is approaching I don't know maybe around negative one but then as we approach from the right looks like the slope is zero so we do not have the same limit of the secant slope as we approach from the left and right hand sides so at both of these points we see the derivative jump and it looks like f of X is not differentiable