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The gradient of the curve $=\frac{a}{x}+bx^2$ at the point (3,6) is 7. Calculate the values of a and b.

I did it,
$6=\frac{a}{3}+b(3^2) \tag{1} $ We also have: (derivative) $y'=-\frac{a}{x^2}+2bx \tag{2}$
$7=-\frac{a}{3^2}+2b(3) \tag{3} $
but it doesn't seen right.

the answer is a=-9, b=1

Can you help me out? thanks.

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    In equation $(1)$ a $b$ is missing.2012-04-13
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    Your values of $a$ and $b$ are right.2012-04-13
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    Why do you think you could be wrong? Anything particular you're doubtful about?2012-04-13
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    yes, i don't know how to do further. thx :)2012-04-13
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    It was not clear if you wanted us to check the answer or how to get that answer? _Oh, well, there are answers now._2012-04-13

2 Answers 2

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From $6=\frac a3+9b$ it follows that $a=18-27b$. Substituting into (3) we get $7=-\frac{18-27b}{9}+6b=3b-2+6b$, hence $b=1$ and $a=18-27=-9$.

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    At the time of this comment, your reputation is $3935$, and mine is $3539$... interesting!2012-04-13
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    @TheChaz A bit strange indeed.2012-04-13
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    how did $a=18-27b$, i got only $18=a+27b$, thx.2012-04-13
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    Sb, if you don't know how to go from $$18 = a + 27b$$ to $$a = 18 -27b$$, then we might have a problem!2012-04-13
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    aww..i see, so sorry, $a+27b=18$ then $a=18-27b$2012-04-13
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    @azarel $7=-\frac{18-27b}{9}$ then $7=3b-2+6b$ but I didn't see 7 in this step in ur answer. thx2012-04-13
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If you multiply equation (3) by three, then add it to equation (1), you'll get $$27 = 27b$$

From this, the value of $a$ follows.