# Calculate the value of a and b.

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.

Can you help me out? thanks.

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In equation $(1)$ a $b$ is missing. – user21436 Apr 13 '12 at 1:34
Your values of $a$ and $b$ are right. – user21436 Apr 13 '12 at 1:36
Why do you think you could be wrong? Anything particular you're doubtful about? – user21436 Apr 13 '12 at 1:38
yes, i don't know how to do further. thx :) – Sb Sangpi Apr 13 '12 at 1:42
It was not clear if you wanted us to check the answer or how to get that answer? Oh, well, there are answers now. – user21436 Apr 13 '12 at 1:44

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! – The Chaz 2.0 Apr 13 '12 at 1:43
@TheChaz A bit strange indeed. – azarel Apr 13 '12 at 1:47
how did $a=18-27b$, i got only $18=a+27b$, thx. – Sb Sangpi Apr 13 '12 at 1:58
Sb, if you don't know how to go from $$18 = a + 27b$$ to $$a = 18 -27b$$, then we might have a problem! – The Chaz 2.0 Apr 13 '12 at 2:01
aww..i see, so sorry, $a+27b=18$ then $a=18-27b$ – Sb Sangpi Apr 13 '12 at 2:05

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.

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