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anonymous

  • 5 years ago

at what points on the curve does y = x^3 slope at 45 degrees?

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  1. anonymous
    • 5 years ago
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    The gradient function is the derivative of your equation. You'd have,\[y=x^3 \rightarrow y'=3x^2\]Now, the slope of a line is equal to the tangent that line makes with the positive horizontal axis. A slope of 45 degrees is then,\[m=\tan 45 ^o=1\]Since y' gives the gradient at any x, you have then,\[y'=1 \rightarrow 3x^2=1\]You then solve for x.

  2. anonymous
    • 5 years ago
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    \[x=\pm \frac{1}{\sqrt{3}}\]

  3. anonymous
    • 5 years ago
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    oh i see now, i didn't realize that a slope of 45 degrees meant tan 45. thankx a million lokisan!

  4. anonymous
    • 5 years ago
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    np :)

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