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Study23

  • 2 years ago

Find an equation to the tangent line to the curve at the given point. (So, I have to find the derivative, but Im stuck - not sure if I am doing this correctly). Function is y=x^(5/2) Point is \(\ (4,32). Please show me step by step! I'm really confused!

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  1. Algebraic!
    • 2 years ago
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    you can do it the usual way.... (x^n) ' = nx^(n-1) n is 5/2

  2. Study23
    • 2 years ago
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    Okay, so that is the derivative....?

  3. Algebraic!
    • 2 years ago
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    yes

  4. Study23
    • 2 years ago
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    I thought it was 5/2(x)^(3/2)

  5. Study23
    • 2 years ago
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    Where do I proceed from here?

  6. Algebraic!
    • 2 years ago
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    plug in x=4 to find the slope (m) of the tangent to the curve at that point.. use y= mx +b with the point they gave in order to find b

  7. Study23
    • 2 years ago
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    Oh Okay. I was trying to plug the derivative in for m, but I guess that isn't correct.

  8. Algebraic!
    • 2 years ago
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    yeah, you just want the value of the derivative at that particular point...

  9. Algebraic!
    • 2 years ago
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    not the general expression for the slope at any value of x..

  10. Study23
    • 2 years ago
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    So Is this the derivative? 5/2(x)^(3/2)

  11. Algebraic!
    • 2 years ago
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    yes

  12. Study23
    • 2 years ago
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    ok thanks for the help!

  13. Algebraic!
    • 2 years ago
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    sure:)

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spraguer (Moderator)
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