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Math Help - Slope of tangent #1

  1. #1
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    Slope of tangent #1

    i) Find the points on the graph of y=(1/3)x^3 -5x-(4/x) at which the slope of the tangent is horizontal.

    ii) find the slope of the demand curve D(p)=20/((sqrt p -1)), p>1, at point (5,10).

    I need help with the two questions above, I'm doing practice questions with the slope of tangent but these last two questions I'm stuck on. Please help me or guide me in the right direction so I can finish these two. Thanks for your help in advance!
    Last edited by mr fantastic; August 12th 2009 at 05:16 PM. Reason: Changed post title
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    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by skeske1234 View Post
    i) Find the points on the graph of y=(1/3)x^3 -5x-(4/x) at which the slope of the tangent is horizontal.

    ii) find the slope of the demand curve D(p)=20/((sqrt p -1)), p>1, at point (5,10).

    I need help with the two questions above, I'm doing practice questions with the slope of tangent but these last two questions I'm stuck on. Please help me or guide me in the right direction so I can finish these two. Thanks for your help in advance!
    1. Take the derivative. Set it to 0. Solve for x. Find the cordinates.

    2. Take the derivative. Evaluate the derivative at x=5.

    Remember: Derivative=Slope=rate of change
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    Ok, for the first question when you say : "1. Take the derivative. Set it to 0. Solve for x. Find the cordinates."

    I am stuck with the first part. When I take the derivative, what do I put in for "a" in the first principle of derivatives formula? ie. [f(a+h) - f(a)]/h

    and then after setting it to 0 and solving for x to find the coordinates.. not sure how to do that part. could you please show me how to begin it and explain more . please and thank you! very much
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    Member McScruffy's Avatar
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    Quote Originally Posted by skeske1234 View Post
    Ok, for the first question when you say : "1. Take the derivative. Set it to 0. Solve for x. Find the cordinates."

    I am stuck with the first part. When I take the derivative, what do I put in for "a" in the first principle of derivatives formula? ie. [f(a+h) - f(a)]/h

    and then after setting it to 0 and solving for x to find the coordinates.. not sure how to do that part. could you please show me how to begin it and explain more . please and thank you! very much
    Do you have to use the limit definition of a derivative (first principle of derivatives) or can you use differentiation rules, in this case the power rule?
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    Quote Originally Posted by McScruffy View Post
    Do you have to use the limit definition of a derivative (first principle of derivatives) or can you use differentiation rules, in this case the power rule?

    i have to use the first principle of derivatives, so how would i go about doing that for the first question?
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    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by skeske1234 View Post
    i have to use the first principle of derivatives, so how would i go about doing that for the first question?
    \frac{d}{dx}\left[\frac{1}{3}x^3-5x-4x^{-1}\right]=(3)\frac{1}{3}x^{3-1}-(1)5x^{1-1}-(-1)4x^{-1-1}=x^2-5+4x^{-2}

    By the power rule.

    If it's first priciples you want do this

    \lim_{h\to0}\frac{f(x+h)-f(x)}{h}

    can you proceed? The answer will be the same as above.

    Ah heck, here ya go. I'll stop bein' lazy.

    \lim_{h\to0}\frac{\overbrace{\left[\frac{1}{3}(x+h)^3-5(x+h)-\frac{4}{(x+h)}\right]}^{f(x+h)}-\overbrace{\left[\frac{1}{3}x^3-5x-\frac{4}{x}\right]}^{f(x)}}{h}
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    Member McScruffy's Avatar
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    Quote Originally Posted by skeske1234 View Post
    i have to use the first principle of derivatives, so how would i go about doing that for the first question?
    \lim_{h\rightarrow0}\frac{\left(\frac{1}{3}(x+h)^3-5(x+h)-\frac{4}{x+h}\right)-\left(\frac{1}{3}x^3-5x-\frac{4}{x}\right)}{h}

    You beat me that time VonNemo19!

    You would in turn work out the derivative, set that equal to zero and solve for x. That is the x-value of the point at which the tangent line is horizontal.
    Last edited by McScruffy; August 2nd 2009 at 06:21 PM.
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    Quote Originally Posted by VonNemo19 View Post
    \frac{d}{dx}\left[\frac{1}{3}x^3-5x-4x^{-1}\right]=(3)\frac{1}{3}x^{3-1}-(1)5x^{1-1}-(-1)4x^{-1-1}=x^2-5+4x^{-2}

    By the power rule.

    If it's first priciples you want do this

    \lim_{h\to0}\frac{f(x+h)-f(x)}{h}

    can you proceed? The answer will be the same as above.

    Ah heck, here ya go. I'll stop bein' lazy.

    \lim_{h\to0}\frac{\overbrace{\left[\frac{1}{3}(x+h)^3-5(x+h)-\frac{4}{(x+h)}\right]}^{f(x+h)}-\overbrace{\left[\frac{1}{3}x^3-5x-\frac{4}{x}\right]}^{f(x)}}{h}

    once i set it to 0 and solve for x, how do i find the coordinates? I have x^4-15x^2-12=0
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  9. #9
    Member McScruffy's Avatar
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    Quote Originally Posted by skeske1234 View Post
    once i set it to 0 and solve for x, how do i find the coordinates? I have x^4-15x^2-12=0
    Plug that x-value back into the original equation to get your y-value. That will give you the (x, y) coordinates of the point where the tangent line is horizontal.
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by McScruffy View Post
    Plug that x-value back into the original equation to get your y-value.
    Indeed, Mcscruffy! Indeed!

    What he said!
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    Quote Originally Posted by VonNemo19 View Post
    Indeed, Mcscruffy! Indeed!

    What he said!

    thanks for your help guys but.. i am still stuck.. i can't seem to find the derivative when I use the [f(x+h)-f(x) ]/h formula.. i cant seem to cancel out the h in the denominator.. could you please show me how to do this? thanks
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    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by skeske1234 View Post
    thanks for your help guys but.. i am still stuck.. i can't seem to find the derivative when I use the [f(x+h)-f(x) ]/h formula.. i cant seem to cancel out the h in the denominator.. could you please show me how to do this? thanks
    Check your algebra one more time, my friend.

    I'll do something similar to show you.

    \lim_{h\to0}\frac{\left[(x+h)^2+\frac{4}{x+h}\right]-\left(x^2+\frac{4}{x}\right)}{h}

    =\lim_{h\to0}\frac{\left[x^2+2hx+h^2+\frac{4}{x+h}\right]-\left(x^2+\frac{4}{x}\right)}{h}

    =\lim_{h\to0}\frac{2hx+h^2+\frac{4}{x+h}-\frac{4}{x}}{h}

    =\lim_{h\to0}\frac{2hx+h^2}{h}+\lim_{h\to0}\frac{\  frac{4}{x+h}-\frac{4}{x}}{h}

    =\lim_{h\to0}\frac{2hx+h^2}{h}+\lim_{h\to0}\frac{\  frac{4x-4(x+h)}{x(x+h)}}{h}

    =\lim_{h\to0}\frac{2hx+h^2}{h}+\lim_{h\to0}\frac{4  x-4x-4h}{h(x)(x+h)}

    Can you take it from here? again... this is not the answer to your problem. It's ver similar though.
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  13. #13
    Member McScruffy's Avatar
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    Quote Originally Posted by skeske1234 View Post
    thanks for your help guys but.. i am still stuck.. i can't seem to find the derivative when I use the [f(x+h)-f(x) ]/h formula.. i cant seem to cancel out the h in the denominator.. could you please show me how to do this? thanks
    Long story short, I wound up with:

    ... (obviously there are a couple of steps before this)

    \lim_{h\to0}\frac{x^4h-5x^2h-5xh^2+4h}{x^2h+xh^2}

    \lim_{h\to0}\frac{x^4-5x^2-5xh+4}{x^2+xh}

    which is: x^2-5+\frac{4}{x^2}, which is the derivative.
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    Quote Originally Posted by VonNemo19 View Post
    Check your algebra one more time, my friend.

    I'll do something similar to show you.

    \lim_{h\to0}\frac{\left[(x+h)^2+\frac{4}{x+h}\right]-\left(x^2+\frac{4}{x}\right)}{h}

    =\lim_{h\to0}\frac{\left[x^2+2hx+h^2+\frac{4}{x+h}\right]-\left(x^2+\frac{4}{x}\right)}{h}

    =\lim_{h\to0}\frac{2hx+h^2+\frac{4}{x+h}-\frac{4}{x}}{h}

    =\lim_{h\to0}\frac{2hx+h^2}{h}+\lim_{h\to0}\frac{\  frac{4}{x+h}-\frac{4}{x}}{h}

    =\lim_{h\to0}\frac{2hx+h^2}{h}+\lim_{h\to0}\frac{\  frac{4x-4(x+h)}{x(x+h)}}{h}

    =\lim_{h\to0}\frac{2hx+h^2}{h}+\lim_{h\to0}\frac{4  x-4x-4h}{h(x)(x+h)}

    Can you take it from here? again... this is not the answer to your problem. It's ver similar though.
    ok, about the fourth line down on yours.. why did you split up the question to two limh>0 + limh>0?
    sorry, i still can't seem to get the h to cancel out on mine.. I will show you what I have
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    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by skeske1234 View Post
    ok, about the fourth line down on yours.. why did you split up the question to two limh>0 + limh>0?
    sorry, i still can't seem to get the h to cancel out on mine.. I will show you what I have
    A strategic move that made my life easier. It is valid because of the theorems on limits says it is.

    I first rewrote the quotient as two separate ones by the basic division identity

    \frac{a+b+c+d}{h}=\frac{(a+b)+(c+d)}{h}=\frac{a+b}  {h}+\frac{c+d}{h}

    And since the limit of a sum is the sum of the limits, I could "distribute" the limit to both quotients. (I probably shouldn't use the word distribute, but who cares)
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