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burhan101

  • 2 years ago

Determine the equation of the tangent to the curve y=

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

  2. burhan101
    • 2 years ago
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    |dw:1371802792014:dw|

  3. oldrin.bataku
    • 2 years 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
    • 2 years ago
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    i think y= \[\huge 2e(x-1)+e\]

  5. Azteck
    • 2 years 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
    • 2 years 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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