Estimate the slope to three decimal places.?

Question

The slope of the tangent line to the graph of the exponential function y=2^x at the point (0,1) is below.

lim (2^(x)-1)/ x Estimate the slope to three decimal places.
x->0

Answer 1

You can use L'Hospital's rule to evaluate
lim (2^x-1)/x
x->0

or note that it is the value of the derivative of y=2^x at x=0. If y=2^x, y'=2^x * ln(2) so

lim (2^x-1)/x=ln(2)
x->0

The value of ln(2) to three decimal places is 0.693

Answer 2

You can use the Taylor series expansion of 2^x:

2^x = 1 + x ln2 + 0.5 (x ln2)^2 + .....

For small x we obtain
(2^x-1)/x = (1 + x ln2 -1)/x = ln 2 = 0.693

Answer 3

Use a table of values.
Start with x = -0.1 and x = 0.1
Then use x = -0.01 and x = 0.01
Keep using smaller and smaller values of x and -x until they have no effect on the third decimal place.

Answer 4

y=2^x

differentiating w.r.t x , y' = {2^x } *{ln 2}

slope of the tangent is y'at (0,1)

that is slope is {2^0 } *{ln 2} = ln(2) =0.693

Answer 5

y=2x is an equation to a straight line. It cannot have any tangent.

Answer 6

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