Attachment 28213Any help with this question would be very appreciated

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- May 1st 2013, 02:56 AMcalmo11Continuity
Attachment 28213Any help with this question would be very appreciated

- May 1st 2013, 03:04 AMProve ItRe: Continuity
Please do not double post.

In order for the function to be continuous at a point, it needs to be defined at that point (which it is at x = 1) and needs to approach the same value from the left as from the right.

So that means in order to be continuous at x = 1, then $\displaystyle \displaystyle \begin{align*} \ln{(1)} = 1.7^1 - C \end{align*}$ (do you see why?) Solve for C. - May 2nd 2013, 12:35 AMcalmo11Re: Continuity
I had done that method, I just didn't get more than 1 value for c, and the question says values, so I was confused. I mainly need help with part b, to show that g is differentiable at x=1.

- May 2nd 2013, 12:39 AMcalmo11Re: Continuity
I am checking by finding the derivatives of both by first principles, and checking if the values that those equations produce at x=1 are the same, I am not too certain whether I am doing the right thing there :S. And I am mainly stuck on the derivative of ln(x) by first principles. I can get to: lim h->0 ln(1+h/x)/h