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Math Help - Rate of change

  1. #1
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    Rate of change

    Consider the function f(x) = 3x^2 - 4x - 1
    a) Estimate the slope of the tangent line at x =1
    b) Find the y-corrdinate of the point of tangency
    c) Use the coordinates of the points of tangency and the slope to find the equation of the tangent line at x = 1

    I'm confused about the slopes and the EQUATION of the line...
    Would the instantaneous rate of change at x = 1 be the slope for a)?
    If it is, then I can use IROC = [ f(x+h) - f(x) ] /h
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  2. #2
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    Quote Originally Posted by TN17 View Post
    Consider the function f(x) = 3x^2 - 4x - 1
    a) Estimate the slope of the tangent line at x =1
    b) Find the y-corrdinate of the point of tangency
    c) Use the coordinates of the points of tangency and the slope to find the equation of the tangent line at x = 1

    I'm confused about the slopes and the EQUATION of the line...
    Would the instantaneous rate of change at x = 1 be the slope for a)?
    If it is, then I can use IROC = [ f(x+h) - f(x) ] /h
    Yes, for a, find the derivative, and plug in 1.

    For b, what is y when x = 1?
    -
    Then use that coordinate in y-y_{1}=m(x-x_{1})
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    Yes, for a, find the derivative, and plug in 1.

    For b, what is y when x = 1?
    -
    Then use that coordinate in y-y_{1}=m(x-x_{1})
    We haven't learned derivatives yet, but here is my solution anyway:
    a) IROC = [ f(x+h) - f(x) ]/ h
    = [ (3(1.01)^2 - 4(1.01) - 1) - (3(1)^2 - 4(1) - 1) ] / 0.01
    = 2.03

    b)The y - coordinate is:
    y = 3(1)^2 - 4(1) - 1
    y = -2

    c) I'm not sure what you mean here..
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  4. #4
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    Quote Originally Posted by TN17 View Post
    We haven't learned derivatives yet, but here is my solution anyway:
    a) IROC = [ f(x+h) - f(x) ]/ h
    = [ (3(1.01)^2 - 4(1.01) - 1) - (3(1)^2 - 4(1) - 1) ] / 0.01
    = 2.03
    It should be 2 not 2.03. Why did you make h = .01?
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    It should be 2 not 2.03. Why did you make h = .01?
    We learned instantaneous rate of change and the teacher chose "0.01" as the h value.
    0.01 is supposed to be the small distance from the given x value on the graph so that we can find the slope of the tangent.
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  7. #7
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    Quote Originally Posted by TN17 View Post
    We learned instantaneous rate of change and the teacher chose "0.01" as the h value.
    0.01 is supposed to be the small distance from the given x value on the graph so that we can find the slope of the tangent.
    \displaystyle\lim_{h\to 0}\frac{3(x+h)^2-4(x+h)-1-[3x^2-4x-1]}{h}\Rightarrow\lim_{h\to 0}\frac{3(1+h)^2-4(1+h)-1-[3-4-1]}{h}=\cdots
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  8. #8
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    Yes use the equation \frac{f(x+h) - f(x) }{h} to estimate the slope. The slope of the tangent line at x=1 IS the rate of change of f(x) at x=1. The
    smaller the value of h you use in the rate of change equation the better the estimate. A thing to note is that for some functions, changing h by a very small amount
    changes that rate of change equation by a large amount so to estimate it accurately enough you would need to use an even smaller value for h. That's where the idea of limits come from and thus leads to differentiation. However, here, the function is a quadratic and using a small number such as h=0.01 is accurate enough. You could also try smaller and smaller values of h such as 0.001 then 0.0001 and you will see that the rate of change function tends to a certain number and then you will see why dwsmith said 2.
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