# max and min point

#### Petrus

##### Well-known member
calculate max and min point of the functuin $\ln(x)+x-2x^2$ in the range $(\frac{1}{6},\frac{3}{2})$
I start to derivate it
$f'(x)=\frac{1}{x}+1-4x$
And my next step is to fined critical point but I got problem to solve this equation.
$\frac{1}{x}+1-4x=0$

#### Jameson

Staff member
Hi Petrus. If you want to solve $$\displaystyle \frac{1}{x}+1-4x=0$$ you can multiply everything by $x$ to get rid of the fraction. The result is $$\displaystyle 1+x-4x^2=0$$. Can you solve it from here?

#### MarkFL

Staff member
calculate max and min point of the functuin $\ln(x)+x-2x^2$ in the range $(\frac{1}{6},\frac{3}{2})$...
Hello Petrus, because you have responded well to $\LaTeX$ tips in the past and use them, I wish to offer you a few here:

Instead of using the \$delimiters, use the$\LaTeX$button on the toolbar which has the $$\displaystyle \sum$$ symbol on it. This will automatically generate the MATH tags for you, which incorporate the \displaystyle command to make your fractions larger. When you wish to enclose tall expressions such as fractions within parentheses (or other bracketing symbols), use: \left( \right) and so the given domain, written as \left(\frac{1}{6},\frac{3}{2} \right) looks like $$\displaystyle \left(\frac{1}{6},\frac{3}{2} \right)$$. Doesn't that look better? #### Petrus ##### Well-known member Hello Petrus, because you have responded well to$\LaTeX$tips in the past and use them, I wish to offer you a few here: Instead of using the \$ delimiters, use the $\LaTeX$ button on the toolbar which has the $$\displaystyle \sum$$ symbol on it. This will automatically generate the MATH tags for you, which incorporate the \displaystyle command to make your fractions larger.

When you wish to enclose tall expressions such as fractions within parentheses (or other bracketing symbols), use:

\left( \right)

and so the given domain, written as \left(\frac{1}{6},\frac{3}{2} \right) looks like $$\displaystyle \left(\frac{1}{6},\frac{3}{2} \right)$$. Doesn't that look better?
Indeed! Thanks Mark I will use it now You could also tell me this early hehe

#### MarkFL

Staff member
It's easier to see what things on which you may benefit from being given tips as they occur, rather than try to think of everything in advance.

Your $\LaTeX$ has come very far in a short period of time...you are doing very well with it!

Have you managed to find the critical points from the derivative?

#### Petrus

##### Well-known member
Hello,
Now I get $$\displaystyle x_1= \frac{1+\sqrt(17)}{8}$$ $$\displaystyle x_2=\frac{1-\sqrt(17)}{8}$$ (When I use that latex toolbar my sqrt does not look right, any tips on how I shall do my sqrt?)
and now im suposed to take second derivate and check if the x wealth is positive or negative ( if its possitive that means its a min point)
$$\displaystyle f''(x)=-\frac{1}{x^2}-4$$
With only typing one of them in calculator I get that $$\displaystyle x_1= \frac{1+\sqrt(17)}{8}$$ is a max point that means also $$\displaystyle x_2=\frac{1-\sqrt(17)}{8}$$ is a min point. I am doing correct so far? and the problem is that my $$\displaystyle x_2=\frac{1-\sqrt(17)}{8}$$ is not in the range? Do i got correct?

#### MarkFL

Staff member
For square roots, use the curly braces, e.g.:

\sqrt{17} gives $$\displaystyle \sqrt{17}$$

This is not because of the toolbar, this is how you should write square roots no matter how you choose to delimit your code.

You have correctly found the roots of the quadratic Jameson helped you to obtain. And you are correct to ignore the root that lies outside the given domain.

So, what you want to do is evaluate the given function (the original, the function you differentiated) at the end-points of the domain AND at the critical value which is within the domain, and take the smallest of these values as the function's absolute minimum and the largest of these as the absolute maximum.

The end-points of the domain and any critical values within that domain are the only places at which absolute extrema may occur for the function on the given domain.

What do you find?

#### Petrus

##### Well-known member
I get that min is $$\displaystyle x=\frac{1}{6}$$ and $$\displaystyle y=\frac{2}{9}-\log(6)$$
and max is $$\displaystyle x=\frac{3}{2}$$ and $$\displaystyle y=\frac{14}{3}+\log(\frac{3}{2})$$

#### MarkFL

Staff member
You are not evaluating the function correctly. Can you show how you arrived at those values for the function at the end-points so we can address what you are doing wrong?

Also, I suggest using your calculator or our Graph Plotter widget to plot the function on the given domain so that you have an idea of what you should find.

#### Petrus

##### Well-known member
You are not evaluating the function correctly. Can you show how you arrived at those values for the function at the end-points so we can address what you are doing wrong?

Also, I suggest using your calculator or our Graph Plotter widget to plot the function on the given domain so that you have an idea of what you should find.
I am not suposed to write all those 3 point on my orginal function $$\displaystyle \ln(x)+x-2x^2$$?

my crit point: ln((1+sqrt(17))/8)+(1+sqrt(17))/8+2((1+sqrt(17))/8)^2 - Wolfram|Alpha
3/2:ln(3/2)+1/6+2(3/2)^2 - Wolfram|Alpha
1/6: ln(1/6)+1/6+2(1/6)^2 - Wolfram|Alpha

#### MarkFL

Staff member
Yes, you should evaluate the function at the end-points and the critical value, but you have made some errors entering them.

In all 3 links you have used $$\displaystyle \ln(x)+x+2x^2$$, while in the 3rd link you have also mistakenly used a mixture of values for $x$. Review and correct your inputs to W|A, then see what you get.

#### Petrus

##### Well-known member
Yes, you should evaluate the function at the end-points and the critical value, but you have made some errors entering them.

In all 3 links you have used $$\displaystyle \ln(x)+x+2x^2$$, while in the 3rd link you have also mistakenly used a mixture of values for $x$. Review and correct your inputs to W|A, then see what you get.
For some reason i still dont get correct...
max: ln((1+sqrt(17))/8)+(1+sqrt(17))/8-2((1+sqrt(17))/8)^2 - Wolfram|Alpha
min: ln(3/2)+3/2-2(3/2)^2 - Wolfram|Alpha

#### MarkFL

Staff member
Those values look right to me. When you say you are not getting them correct, do you mean when you enter the values into your homework software they are being rejected?

Are you supposed to use decimal approximations, and if so how much precision is required? Please tell me the exact instructions you are given regarding entering your results.

#### Petrus

##### Well-known member
Those values look right to me. When you say you are not getting them correct, do you mean when you enter the values into your homework software they are being rejected?

Are you supposed to use decimal approximations, and if so how much precision is required? Please tell me the exact instructions you are given regarding entering your results.
Thanks Mark and Jameson!
You both helped me and Mark your guide was so great!I do understand now!
(I did input x and y wealth that was wrong, I was suposed to only input y wealth)

#### MarkFL

Staff member
You are so good about giving feedback and letting those who help you know how things turn out!

By the way, you want to use the word "value" not "wealth." While the two words are closely related, in English wealth is not used to denote the magnitude of a numeric quantity (unless it is monetary).

#### Petrus

##### Well-known member
You are so good about giving feedback and letting those who help you know how things turn out!

By the way, you want to use the word "value" not "wealth." While the two words are closely related, in English wealth is not used to denote the magnitude of a numeric quantity.
Thanks for correcting me on this thing Indeed I need to improve my english cause if it works good for me ima try studdy outside Sweden Btw do you say calculate the max and min point or do you say Find the max and min point?