I already found $\frac{dy}{dx}$ to be $9-\frac{1}{x^2}$, however I don't know what $x$ needs to be for $\frac{dy}{dx}$ to equal zero, could someone please explain how they get to the answer for $x$?
Find co-ordinates of point on curve where the gradient is 0
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calculus
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09$x$ - 2 + $\frac{1}{x}$ – 2017-01-09
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0It is a trivial quadratic equation. https://en.wikipedia.org/wiki/Quadratic_equation – 2017-01-09
2 Answers
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If the derivative is given by $9-\frac{1}{x^2}$, you only need to set this equal to 0 and solve for $x$:
$$9-\frac{1}{x^2}=0$$
$$9=\frac{1}{x^2}$$
$$x^2=\frac{1}{9}$$
$$x=\pm \frac{1}{3}$$
It doesn't matter what the original function is ($y(x) = 9x+\frac{1}{x}+C$ where $C$ is a constant), that will only determine the value of the function at these points.
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Here is a graphical answer to your question. The curve given by $9 -\frac{1}{x^2}$ equals zero at these two points.
