Need help

[shahnaz]

need help with this problem !

 

[Deleted member 4993]

need help with this problem !

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[Dr.Peterson]

Though you didn't show your work or thinking, you did show your answers, which are revealing.

You were told to compute the derivative; the limits mentioned are not limits of f(x), which you have done, but of the difference quotient!

And then you seem to be supposing that the derivative is the sum of the limits on each side. Why? Or did you have some other way to get that correct derivative?

This is all about the definition of the derivative; looking at that definition probably would have kept you from going in the wrong direction.

 

[HallsofIvy]

The limits of $\displaystyle 6x- 9x^2$ and $\displaystyle 3x^2+ 6x$ are 0, not 3!
But, as Dr. Peterson said, you were not supposed to take the limits of those but of their derivatives, $\displaystyle 6- 18x$ and $\displaystyle 6x+ 6$ which are both 6. And the derivative at x= 0 is that common value, not their sum.

 

[Steven G]

Seriously, you need to rethink about your 0 table. 0*0 = 0, 0*1=0, 0*2=0, 0*3=0,... Do you see a pattern here.

 

[Steven G]

I disagree with the previous two posters. We all know that if f'(0) exists, then the function must be continuous at x=0. Sure the left hand and right hand derivative may exist and might even be equal, but equality is not enough to conclude that the derivative exists. You must also confirm that the function is continuous at x=0 (which it is).

 

[Steven G]

And the derivative at x= 0 is that common value.

I think that you are guilty of sloppy thinking here. We all do that once in a while!

 

[Dr.Peterson]

The limits of $\displaystyle 6x- 9x^2$ and $\displaystyle 3x^2+ 6x$ are 0, not 3!
But, as Dr. Peterson said, you were not supposed to take the limits of those but of their derivatives, $\displaystyle 6- 18x$ and $\displaystyle 6x+ 6$ which are both 6. And the derivative at x= 0 is that common value, not their sum.

Actually, what I said was to take the limit of the difference quotient, not of the derivative (though I'd thought of the latter, too). The problem explicitly says to use the definition of the derivative.

So we take the limits of (6x - 9x^2)/x and of (3x^2 + 6x)/x, which are both 6.