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for differentiability Lf'(x)=Rf'(x)
Here Lf'(3)=2.3=6 and Rf'(3)=m so we must have ,m=6
Where does LF come from?
Lf'(x) means the left hand derivative of f at the point x
and Rf'(x) means the right hand derivative of f at the point x
so for mx+b you would plug in 3 for x
Wait... yeh ignore that. Sorry. Im super confused.
when x<=3, f(x)=x^2 and f'(x)=2x
when x>3, f(x)=mx+b and f'(x)=m
Oh the derivatives of each. Okay i see now. right,
Lf'(3)=2.3=6 and Rf'(3)=m so both these must be equal for the function to be differentiable at x=3, so 6=m
and here b can take any real value. it has no restriction
Okay I understand that now, except for the b. Finding b. How is it any real value if its being added to m. and m is 6.
the differentiability has no relation with b here.. since f'(x) doesnot contain any thing related to b. so b can be any value,
Okay. I think I have to find an actual value of b.
there will be no actual value. you take any place and place it for instance
oh wait a bit i think i have the answer
see the function will be differentiable, so it will be continuous as well
now since we know the value of m, let us re-write the function
6x+b but b=-9 so its 6x-9 when x >3
f(x)=x^2,x<=3 f(x)=6x+b,x>3 now the left hand limits at x=3 is 3^2=9 and the right hand limit at x=3 is 18+b so 9=18+b so b=-9
And x^2 and 6x-9 connect together when graphing.
Good, we both got it now. ahaha.
hence we have m=6,b=-9