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anonymous

  • 4 years ago

the derivate of h(x)=4/7x is 4/7?

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  1. anonymous
    • 4 years ago
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    yup

  2. anonymous
    • 4 years ago
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    ok i got it now

  3. anonymous
    • 4 years ago
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    jimmy can you help me with this find the slope of the tangent line of f(x)= 2x^2-x+3

  4. anonymous
    • 4 years ago
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    best written as (6/7) x though jinnie - so its not confused with 4 divided by 7x

  5. anonymous
    • 4 years ago
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    6/7?

  6. anonymous
    • 4 years ago
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    yea ok find derivative of f(x) f'(x) = 4x - 1 - thats it

  7. anonymous
    • 4 years ago
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    so for example the slope of tangent at x = 1 = 4(1) - 1 = 3

  8. anonymous
    • 4 years ago
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    sorry i meant (4/7)

  9. anonymous
    • 4 years ago
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    so i must find the derivative first which is 2x-1+0

  10. anonymous
    • 4 years ago
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    of 2x^2-x+3

  11. anonymous
    • 4 years ago
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    yeah - but its 4x - 1 The derivative at a point on a curve is slope of the tangent at that point

  12. anonymous
    • 4 years ago
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    oh yea you have to multiply ok i got it

  13. anonymous
    • 4 years ago
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    so what is the equation of the tangent line at x=1? is it the same as the slope of the tangent line at the point x=1?

  14. anonymous
    • 4 years ago
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    no use the formula y-y1 = m(x-x1) to get the equation at x=1 , y = 2(1)^2 - 1 + 3 = 4 so we need the equation at the point (1,4) with a slope of 3 y - 4 = 3 (x - 1) is equation of the tangent

  15. anonymous
    • 4 years ago
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    can i continue with that or is that where it stops

  16. anonymous
    • 4 years ago
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    you can simplify that y - 4 = 3x - 3 y = 3x + 1

  17. anonymous
    • 4 years ago
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    so the way to find the equation of the tangent at a point on a curve is to 1. differentiate the equation of the curve to find slope of the curve 2 substitute for slope, and x and y values of the point into the general form y-y1 = m(x-x1)

  18. anonymous
    • 4 years ago
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    alright thanks

  19. anonymous
    • 4 years ago
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    how did you get the slope of 3 btw

  20. anonymous
    • 4 years ago
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    yw - hope this helps

  21. anonymous
    • 4 years ago
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    i found the slope in terms of x by differentiating f(x) = 2x^2 - x +3 f'(x) = 4x - 1 - this the slope in terms of x , at point x=x so at x = 1, slope is found by plugging x=1 into 4x-1: slope = 4(1) - 1 = 3

  22. anonymous
    • 4 years ago
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    ok thanks

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