How Do You Estimate the Slope of a Tangent to y=2^x at (0,1)?

  • Thread starter Thread starter fstam2
  • Start date Start date
  • Tags Tags
    Limits Slope
fstam2
Messages
10
Reaction score
0
Hey there, I need a little push in the right direction.
Here is the question:
The slope of the tangent line to the graph of the exponential function
y=2^x at the point (0,1) is lim x approaches 0 (2^x-1)/x.
Estimate the slope to three decimal places.
Where I am getting confused is which formula do I plug (x) into to find a secant slope?
I hope I asked the right question.
Thanks
 
Physics news on Phys.org
fstam2 said:
Hey there, I need a little push in the right direction.
Here is the question:
The slope of the tangent line to the graph of the exponential function
y=2^x at the point (0,1) is lim x approaches 0 (2^x-1)/x.
Estimate the slope to three decimal places.
Where I am getting confused is which formula do I plug (x) into to find a secant slope?
I hope I asked the right question.
Thanks

Yes, since you are only asked to "estimate the slope to three decimal places" you just need to plug a small enough x into that secant formula which is exactly what you wrote: (2x-1)/x. That should take about 5 seconds using a calculator!

(Finding the limit itself in order to find the actual formula is much harder!)
 
Thank you for your help
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top