I need to find the function $f(x)$ that is tangent to a line whose slope is given by $\displaystyle \frac{(1+\sqrt x)^{\frac{1}{2}}}{8\sqrt x}$ that passes through the point $(9,8/9)$.

I really don't know what to do with this. I tried finding the limit but that didn't help and I couldn't do the complex nature of it. I really hope someone can help give me some clarity on what to do with this problem.

  • 1
    $\begingroup$ The slope of the tangent at some point $(x,f(x))$ equals $f'(x)$, i.e $f'(x)=\dfrac{(1+\sqrt{x})^{\frac{1}{2}}}{8\sqrt{x}}$. Do you know how to integrate $f'(x)$? If you do, than integrate and then use the given point in order to find the integration constant. Good luck! $\endgroup$ – Galc127 Jun 14 '16 at 16:53

The slope of a tangent to your function is given by:

$$f'(x)=\displaystyle \frac{(1+\sqrt x)^{\frac{1}{2}}}{8\sqrt x}$$

Thus to retreat the function we integrate:

$$f(x)=\int \displaystyle \frac{(1+\sqrt x)^{\frac{1}{2}}}{8\sqrt x} dx$$

Try the substitution $u=1+\sqrt{x}$

You'll get $f(x)=g(x)+c$

Plug in $x=9$, $y=f(x)=\frac{8}{9}$ and solve for $c$

$$ \int \displaystyle \frac{(1+\sqrt x)^{\frac{1}{2}}}{8\sqrt x} dx=\int \frac{u^{1/2}}{4} du=\frac{u^{3/2}}{6}+c=\frac{(1+\sqrt{x})^{3/2}}{6}+c $$ $$\frac{8}{9}=\frac{(1+\sqrt{9})^{3/2}}{6}+c$$ $$\frac{8}{9}=\frac{8}{6}+c$$ $$\frac{8}{9}=\frac{4}{3}+c$$ $$\frac{8}{9}=\frac{12}{9}+c$$ $$-\frac{4}{9}=c$$ $$f(x)=\frac{(1+\sqrt{x})^{3/2}}{6}-\frac{4}{9}$$

  • $\begingroup$ in your final answer you lost the (-) sign on the C so its actually -4/9 $\endgroup$ – google Jun 14 '16 at 17:31

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