Find the equation of the tangent line to the curve $f(x)=\sqrt{x^3}\ln(x)$ that is parallel to the x-axis.

Question

I have gotten up to $$f'(x)=\dfrac{3\sqrt{x}\ln(x)}{2} + \sqrt{x}$$

I am assuming $m=0$ because when I plug $0$ in the derived function it gives me $0$. I am now not too sure on how to find a tangent line that is parallel to the x-axis.

Answer 1

Your derivative looks fine. You need to factor out the square root and inside the parenthesis you get $\frac{3}{2}lnx+1$. It's that term you need to set equal to zero. This comes to $lnx=-\frac{2}{3}$ from which you can find $x$. Since $0$ is not in the domain of $f(x)$, this value does not count. You may make a graph to see what is going on, and observe there is an (absolute) minimum in the fourth quadrant. That's the point you need

Answer 2

You are solving

$$0=f'(x)=\dfrac{3\sqrt{x}\ln(x)}{2} + \sqrt{x}$$

Setting $y=\sqrt{x}$ gives

$$0=\frac{3}{2}y\log(y^2)+y=3y\ln(y) + y$$

$$y=-3y\log(y)$$

$$-\frac{1}{3}=\log(y)$$

$$y=e^{-1/3}$$

$$x=e^{-2/3}$$

Then the tangent line is the horizontal line

$$0+\sqrt{(e^{-2/3})^3}\ln(e^{-2/3})=-\frac{2}{3 e}$$