to be differentiable it has to be continuous so \[x^2 = mx+b~at~x = 2\] and the slopes have to be the same so \[2x = m~at~x=2\]
plug in 2 and then solve the system for m and b
Would I plug it into x^2=mx and solve for m?
you're eventually going to have to put it into both. if you start with 2x = m, then you can get m and substitute it into the top one and solve for b
So m would equal 4 and b would equal -4?
And thats the answer?
Okay thank you!!!