# Thread: Given a function, find points, find limits, etc.

1. ## Given a function, find points, find limits, etc.

Given the function $\displaystyle f(x) = x^2ln(x)$, x is contained in the set (0,1)

A) Find the coordinates of any points where the graph of f has a horizontal tangent line.

B) Find the coordinates of any points of inflection on graph f

C) Find lim as $\displaystyle x --> 0^+ f(x)$ AND lim as $\displaystyle x --> 0^+ f'(x)$

D) Sketch graph of f using info obtained in a,b,c, with coordinates, max, min, etc.

The first 2 parts are based on units I did a WHILE ago ...

Do I do this for A?

$\displaystyle dy/dx = x + 2xln(x)$
$\displaystyle 0 = x + 2xln(x)$ (because 0 = horizontal line)

And solve for x?

2. Do I do this for A?

(because 0 = horizontal line)

And solve for x?

absolutely

3. Originally Posted by Lord Darkin
Given the function $\displaystyle f(x) = x^2ln(x)$, x is contained in the set (0,1)

A) Find the coordinates of any points where the graph of f has a horizontal tangent line.

B) Find the coordinates of any points of inflection on graph f

C) Find lim as $\displaystyle x --> 0^+ f(x)$ AND lim as $\displaystyle x --> 0^+ f'(x)$

D) Sketch graph of f using info obtained in a,b,c, with coordinates, max, min, etc.

The first 2 parts are based on units I did a WHILE ago ...

Do I do this for A?

$\displaystyle dy/dx = x + 2xln(x)$
$\displaystyle 0 = x + 2xln(x)$ (because 0 = horizontal line)

And solve for x?
FoR C)

$\displaystyle \lim_{x\to0^+}x^2\ln{x}=\lim_{x\to0^+}\frac{\ln{x} }{x^{-2}}=\lim_{x\to0^+}\frac{x^{-1}}{-2x^{-3}}$ by L'Hopital's rule.

4. Part a, my math solver on the calculator says -1.7632 (I had to use the absolute value of x to put in for the natural log).

But I'm confused here since x is supposed to be contained in the set from 0 to 1.

VonNemo19, thanks for the help, I'll look into that in more depth once I understand part a and b more.

5. you have x + xln(x) = 0

x(1+2ln(x)) = 0

x= 0 or e^(-1/2) = .606

0 is not in (0,1) so e^(-1/2) is the only pt on the interval with a horizontal tangent

6. $\displaystyle y = x^2 \cdot \ln{x}$

$\displaystyle y' = x + 2x\ln{x} = x(1 + 2\ln{x}) = 0$

reject $\displaystyle x = 0$ as a solution since $\displaystyle x \in (0,1)$

$\displaystyle \ln{x} = -\frac{1}{2}$

$\displaystyle x = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}}$

7. Ahh, I get it.

So how does everything look?

Part A

x = 0.6065

y = -0.184 (Makes sense? - plugged it back in f(x))

Part B

$\displaystyle y'' = 3 + 2lnx$
$\displaystyle 0 = 3 + 2lnx$
$\displaystyle x = e^{-\frac{3}{2}}$
$\displaystyle x = 0.2231$
$\displaystyle y = -0.446$ (Plugged back in f ' (x) )

Part C

$\displaystyle \lim_{x\to0^+}[x^2\ln{x}] = \lim_{x\to0^+}\frac{\ln{x}}{x^{-2}} = \lim_{x\to0^+}\frac{x^{-1}}{-2x^{-3}} = 0$

$\displaystyle \lim_{x\to0^+}[x + 2xlnx] = (\lim_{x\to0^+}[x] + \lim_{x\to0^+}[2xlnx]) = \lim_{x\to0^+} -2x = 0$

Part D is just graphing, I should be fine with that.

8. looks great.

however $\displaystyle \lim_{x\to0^+}2x\ln{x}$ yields an inderminate form. rewrite as a quotient and Apply L'Hopital's rule.

9. ^yes, I used L'Hopital's rule, but didn't show it there (takes a while for me to type the latex). It came out to be -2x.

The only weird thing is, when I graph this on my calc, the point of inflection for part b doesn't make sense since the y value is -0.446 in part b but the graph on my calc is more like -0.2.

10. Originally Posted by Lord Darkin
^yes, I used L'Hopital's rule, but didn't show it there (takes a while for me to type the latex). It came out to be -2x.

The only weird thing is, when I graph this on my calc, the point of inflection for part b doesn't make sense since the y value is -0.446 in part b but the graph on my calc is more like -0.2.
Hint:
put $\displaystyle x=\frac{1}{t}$.
$\displaystyle t\to \infty ,x\to 0$
now find,
$\displaystyle \lim_{t\to \infty }-2\left (\frac{\ln t}{t} \right )$

11. Originally Posted by Lord Darkin
^yes, I used L'Hopital's rule, but didn't show it there (takes a while for me to type the latex). It came out to be -2x.

The only weird thing is, when I graph this on my calc, the point of inflection for part b doesn't make sense since the y value is -0.446 in part b but the graph on my calc is more like -0.2.
$\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}[x+2x\ln{x}]=1+(2x\frac{1}{x}+2\ln{x})=3+2\ln{x}=0\Rightarrow\ ln{x}=-\frac{3}{2}\Rightarrow{x}=e^{-3/2}$

12. Originally Posted by Raoh
Hint:
put $\displaystyle x=\frac{1}{t}$.
$\displaystyle t\to \infty ,x\to 0$
now find,
$\displaystyle \lim_{t\to \infty }-2\left (\frac{\ln t}{t} \right )$

What is the purpose of doing that? I'm just wondering since it seems like I still get lim f '(X) = 0.

Also, I understand that e=^(-3/2) but I'm confused about the y value for the point of inflection. My calculator shows that the lowest point of the graph of f(x) is -0.183 so why does the point of inflection I have in part b have y=-0.446?

13. Originally Posted by Lord Darkin
What is the purpose of doing that? I'm just wondering since it seems like I still get lim f '(X) = 0.

Also, I understand that e=^(-3/2) but I'm confused about the y value for the point of inflection. My calculator shows that the lowest point of the graph of f(x) is -0.183 so why does the point of inflection I have in part b have y=-0.446?
$\displaystyle f(e^{(-3/2)})\approx{-.075}\neq-.446$

14. Originally Posted by Lord Darkin
What is the purpose of doing that? I'm just wondering since it seems like I still get lim f '(X) = 0.

Also, I understand that e=^(-3/2) but I'm confused about the y value for the point of inflection. My calculator shows that the lowest point of the graph of f(x) is -0.183 so why does the point of inflection I have in part b have y=-0.446?
sorry about that i thought that would help (Post 8).

15. Oh!!! I substituted in the x value that I got in part b into the first derivative, not the original!!!

Need to remember that.

Thanks everyone!

Problem Solved!