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.
$\endgroup$ 52 Answers
$\begingroup$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 $a=18-27=-9$.
$\endgroup$ 6 $\begingroup$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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