The question I'm working on is, "Find the approximate value of the slope of the tangent to the following functions at the given x-value. Use the algebraic method." The first question is y=1/x at x=1. Just plug in 1 for x in the equation right? By doing that I would have the approximate value of the slope correct?
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f(x)=1/x
f '(x)=-1/x^2
f '(1)=-1/1
f '(1)=-1
f '(x)=-1/x^2
f '(1)=-1/1
f '(1)=-1
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One way to approximate the slope of the tangent line is to use the fact that the tangent line at a point (x, f(x)) is the limit of secant lines that pass through (x, f(x)) and (x + h, f(x + h)) as h --> 0. Since we want to approximate this slope, we don't need to compute the actual limit; instead, we can pick values of h sufficiently close to 0 to get our approximation.
Any point on the curve y = 1/x can be represented as (x, 1/x). Then, the secant line that passes through (1, 1) and (1 + h, 1/(1 + h)) has slope:
m = ∆y/∆x = [1/(1 + h) - 1/1]/[(1 + h) - 1] = [1/(1 + h) - 1]/h.
If you take the limit of this as h --> 0, you will get the exact slope. If you plug in a number for h close to 0, like 0.01, you will get a decently accurate value for the slope. With h = 0.01, we see that:
[1/(1 + h) - 1]/h = [1/(1 + 0.01) - 1]/0.01 ≈ -0.990099,
which is close to the exact value of the slope of the tangent line (-1).
I hope this helps!
Any point on the curve y = 1/x can be represented as (x, 1/x). Then, the secant line that passes through (1, 1) and (1 + h, 1/(1 + h)) has slope:
m = ∆y/∆x = [1/(1 + h) - 1/1]/[(1 + h) - 1] = [1/(1 + h) - 1]/h.
If you take the limit of this as h --> 0, you will get the exact slope. If you plug in a number for h close to 0, like 0.01, you will get a decently accurate value for the slope. With h = 0.01, we see that:
[1/(1 + h) - 1]/h = [1/(1 + 0.01) - 1]/0.01 ≈ -0.990099,
which is close to the exact value of the slope of the tangent line (-1).
I hope this helps!
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Slope = dy/dx = -1/x^2
When x = 1 slope = -1
When x = 1 slope = -1
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-1