How do you find the equation of the line tangent to #y=2^x# that passes through the point (1,0)?
That is:
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To find the equation of the line tangent to y=2^x that passes through the point (1,0), we need to find the slope of the tangent line at the point (1,0) and then use the point-slope form of a linear equation.
The slope of the tangent line can be found by taking the derivative of the function y=2^x with respect to x. The derivative of 2^x is ln(2) * 2^x.
Substituting x=1 into the derivative, we get ln(2) * 2^1 = 2ln(2).
So, the slope of the tangent line at the point (1,0) is 2ln(2).
Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute the values to find the equation of the tangent line.
Substituting (1,0) and 2ln(2) into the equation, we get y - 0 = 2ln(2)(x - 1).
Simplifying, the equation of the line tangent to y=2^x that passes through the point (1,0) is y = 2ln(2)(x - 1).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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