I'm trying to find the function $f(x)$ whose tangent line has the slope

$\frac{(1+\sqrt x)^{1/2}}{8\sqrt x}$ for any $x\neq 0$ and whose graph passes through the point $(9,\frac{3}{4})$.

I'm not sure how to even start this problem.

  • 1
    $\begingroup$ Saying that the tangent line of f(x) has the slope ((1+√x)^(1/2))/(8√x) is exactly the same as saying that d/dx f(x) = ((1+√x)^(1/2))/(8√x). ie. f(x) = integral(((1+√x)^(1/2))/(8√x)) + C .. solve the integral and plug in the values f(9)=3/4 to find C $\endgroup$ Commented Jan 10, 2019 at 21:42
  • $\begingroup$ The slope of the tangent to a function is given by the derivative of that function. So you are given the derivative of the function and you are asked to find the function itself. How do you do that? $\endgroup$
    – NickD
    Commented Jan 10, 2019 at 21:42

3 Answers 3


Welcome to Math Stack Exchange. I think that maybe the best way to proceed to write down what we know!

We know what tangent line means, it means derivative. That is,


and if the graph of the function passes through $(9,3/4)$. This would be the same as saying:$ f(9)=3/4$

Now, from the Fundamental Theorem of Calculus we can integrate $f'(x)$ to get $f(x)$. And then using the point that they gave us we can solve for the constant of integration. Do you know how to proceed from here? Do you know how to integrate that?



We have that $$f'(x)={\sqrt{1+\sqrt x}\over 8\sqrt x}$$since $x>0$ by substituting $x=u^2$ we obtain$$f'(u^2)={\sqrt {1+u}\over 8u}$$or equivalently$$2uf'(u^2)={\sqrt{1+u}\over 4}$$ by integrating the sides we conclude that $$f(u^2)={1\over 6}(1+u)\sqrt{1+u}+C$$


You’re given the function of the slope of the tangent of some function $f(x)$. In other words, you’re given the derivative, or $f’(x)$. You reach $f’(x)$ through differentiating.

In the reverse process, you reach $f(x)$ through integrating:

$$f’(x) = \frac{\left(1+\sqrt{x}\right)^{\frac{1}{2}}}{8\sqrt{x}}$$

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

As a hint for evaluating the integral, $\frac{d}{dx}\left(1+\sqrt{x}\right) = \frac{1}{2\sqrt{x}}$, so you may want to use substitution.

Finally, after integrating, you’re given a point the function passes through: $\left(9, \frac{3}{4}\right)$. Recall that when integrating, you also need to add a constant $C$. Once you integrate, plug in $y = 9$ and $x = \frac{3}{4}$ and find what the constant value is. Then, write the function $f(x)$ with the obtained value and you’re done.


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