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I've got a function $f(x) = 3(ax - \frac{b}{x})^3$ that I need to find a and b.

I'm given the information:

$f(\frac{3}{2}) = 3$ and $f'(\frac{3}{2}) = 30$

I've calculated the derivative to be the following

$f'(x) = 9(ax - \frac{b}{x})^2 (ax + \frac{b}{x^2})$

I know the answers are a = 2 and b = 3 but can't workout how to get them using the derivative. Can someone please shed some light on this?

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    Write down everything you can in your question. Substitute the values $x = \frac 32$ in both equations, and write down the resulting equations in terms of $a$ and $b$. – 2017-02-14
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    Take care : as typed, the derivative is wrong. – 2017-02-14
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    Sorry I was tired when I typed it up, it shouldn't have the x term with ax – 2017-02-15

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we have $$f(x)=3\left(ax-\frac{b}{x}\right)^3$$ and we get $$f\left(\frac{3}{2}\right)=3\left(a\frac{3}{2}-\frac{2b}{3}\right)^3=3$$ and $$f'\left(\frac{3}{2}\right)=9\left(ax-\frac{b}{x}\right)^2\left(a+\frac{b}{x^2}\right)$$ and we get $$f'\left(\frac{3}{2}\right)=9\left(a\frac{3}{2}-\frac{2}{3}b\right)^2\left(a+\frac{4}{9}b\right)=30$$ you must solve These two equations for $a$ and $b$

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    What does the beginning of solving this look like? I end up with $6 = 9a-4b$ for the initial equation and $(9a+4b)(9a-4b)^2 = 1080$ which ends up pretty messy when expanding it out. What am I missing here? I got a feeling it's something simple but I'm unsure – 2017-02-15
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    Never mind so more time spent on it with a clear head has solved it. Thanks! – 2017-02-15