the sl0pe of the tangent line to the graph of the exponential function y=2^x at the point (0,1) is
lim(x->0)(2^x-1)/x.
estimate the slope of three decimal places.
f'(x)=2^x. what do i do. we have to solve this without using derivative. i keep getting zero.
you have the derivative wrong.
f'(x)= 2^x * ln2
But you cant use derivatives...Can you use L'Hopitals Rule for limits of the form 0/0 ?
If so, then the limit becomes
slope= (2^x ln2 -0)/1 and as x>0, this is ln 2.
To estimate the slope of the tangent line to the graph of the exponential function y=2^x at the point (0,1) without using the derivative, we can use the given expression:
lim(x->0)(2^x-1)/x
Let's evaluate this limit step by step:
1. Substitute the value of x into the expression:
(2^0-1)/0
2. Simplify the numerator:
(1-1)/0
3. Simplify the fraction:
0/0
At this point, we have encountered an indeterminate form, which means further analysis is needed to find the limit.
To solve this, we can use L'Hopital's Rule, which states that if the limit of a fraction of two functions in an indeterminate form is taken, the limit equals the limit of the derivative of the numerator divided by the derivative of the denominator.
Let's apply L'Hopital's Rule to the expression:
lim(x->0)(2^x-1)/x = lim(x->0)(d/dx(2^x-1))/(d/dx(x))
Now, let's find the derivatives of the numerator and the denominator:
Derivative of the numerator: d/dx(2^x-1) = ln(2) * 2^x
Derivative of the denominator: d/dx(x) = 1
Replace the derivatives back into the expression:
lim(x->0)(ln(2) * 2^x)/(1)
Since the derivative of x is simply 1, we can simplify the expression further:
lim(x->0)(ln(2) * 2^x)
Now, let's evaluate the limit numerically. Plug in x=0 into the expression:
(ln(2) * 2^0) = (ln(2) * 1) = ln(2)
Therefore, the estimated slope of the tangent line to the graph of the exponential function y=2^x at the point (0,1) is ln(2) (approximately 0.693).
To find the slope of the tangent line to the graph of the exponential function y=2^x at the point (0,1) without using the derivative directly, you can use the limit definition of the derivative.
Let's apply the limit definition: lim(x->0)(2^x - 1)/x.
To evaluate this limit, you can simplify the expression by using algebraic manipulations.
Step 1: Substitute x=0 into the expression.
lim(x->0)(2^x - 1)/x = (2^0 - 1)/0 = (1 - 1)/0 = 0/0.
At this point, we have an indeterminate form (0/0), which does not provide a definitive answer.
Using L'Hôpital's Rule, we can differentiate the numerator and denominator separately.
Step 2: Differentiate the numerator: d(2^x - 1)/dx = ln(2) * 2^x.
Step 3: Differentiate the denominator: d(x)/dx = 1.
Now, let's evaluate the limit again using the differentiated expressions.
lim(x->0)(2^x - 1)/x = lim(x->0)(ln(2) * 2^x)/1.
Step 4: Substitute x=0 into the differentiated expressions.
lim(x->0)(ln(2) * 2^x)/1 = ln(2) * 2^0/1 = ln(2)/1 = ln(2).
Therefore, the slope of the tangent line to the graph of y=2^x at the point (0,1) is ln(2).
As for your calculation, it seems that you obtained a result of zero because you mistakenly evaluated the numerator as 2^0 - 1 = 1 - 1 = 0. However, this is incorrect. You should have obtained ln(2) using the steps described above.