For Instance my graph would be x²

lets say the intervals are from [o,3] so then a=0 and b=3 now we got the points now all we need is to use them in the equation and find c so F(3)-F(0)/3-0 =3 so the slope of the secant line is 3 and the equation is the derivative 2x

Ok so skipping all of the equations blah blah blah we find that c lies on x=1 and knowing that we can do the tangent line. Which again blah blah blah we find that it passes through (1,1)and since the tangent line is parallel to the secant use that knowledge and pont slope form and then we get the equation F(x)=2x-1 and bam definition graphically

2. this doesnt work for equations that end up being cusps, corners, etc like and absolute value equation it is not diferrentible at x=0 so there is no tangent point there right? hehehe

"this POINT of Secant line is equal to the POINT of the Tangent line."
ReplyDeleteWhoa whoa. I think you're missing a VERY important word here...
Yes, the corner at x=0 makes this graph non differentiable, but why does this mean I can't do the Mean Value Theorem for it?
can you be a little more specific please. about the second one. i'm having trouble trying explain the second problem. hahahaha.
ReplyDeletewhy does it not work at x=0???
Can you develop your second question a bit more? Try adding boundaries, and drawing some tangent and secant lines, even if one doesn't exist.
ReplyDeleteand try to be more specific so it can be clearer (:
isnt it hard to explain this stuff?
ReplyDeletea little more explanation of your graphs and it will help alot
ReplyDeleteexplanation on your 2nd :D
ReplyDelete