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burhan101

  • one year ago

Determine the equation of the tangent to the curve y=

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  1. burhan101
    • one year ago
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    \[\huge y=e^{2x-1} \] at x=1

  2. burhan101
    • one year ago
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    |dw:1371802792014:dw|

  3. oldrin.bataku
    • one year ago
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    our tangent passes through point \((1,e)\) with slope determined by:$$y'=e^{2x-1}\cdot2=2e^{2x-1}\\y'(1)=2e$$thus we just put it all together:$$y-e=2e(x-1)\\y=2ex-2e+e=2ex-e$$

  4. burhan101
    • one year ago
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    i think y= \[\huge 2e(x-1)+e\]

  5. Azteck
    • one year ago
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    @burhan101 When they give you the x value of the point, you should start finding the y-value for the point by plugging the x-value which is 1, into the curve's equation. Once you've done that and found the derivative (which you've done), you can find the gradient of the tangent by plugging the point into the derivative.

  6. Azteck
    • one year ago
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    Then you can use this to find your equation of the tangent: \[x-x_{1}=m(y-y_{1})\] where: \[(x_{1}, y_{1})\] is the point given and m is the gradient of the tangent

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