# Thread: Help with calculus problem

1. ## Help with calculus problem

Hi. Someone can help me with this.

$f(x)=x-ln x$

a) Discover in the grafic of the funcion the point with abscissa c knowing that the line that is tangent with c is parallel with the secant line that passes on the points with abscissa 1 and e.

b) Show that $x \ge 1+ ln x$

2. Originally Posted by usual_suspect
Hi. Someone can help me with this.

$f(x)=x-ln x$

a) Discover in the grafic of the funcion the point with abscissa c knowing that the line that is tangent with c is parallel with the secant line that passes on the points with abscissa 1 and e.
what does "abscissa" mean?

b) Show that $x \ge 1+ ln x$
consider the function $f(x) = x - \ln x - 1$ defined on $(0, \infty)$

show that this function is always nonnegative. u can do this by showing it is nonnegative on (0,1] then show it is strictly increasing for x > 1

3. Originally Posted by Jhevon
what does "abscissa" mean?
abcissa is the x-coordinate or the coordinate in the x-axis
Cartesian coordinate system - Wikipedia, the free encyclopedia
Thanks

4. Originally Posted by Jhevon
consider the function $f(x) = x - \ln x - 1$ defined on $(0, \infty)$

show that this function is always nonnegative. u can do this by showing it is nonnegative on (0,1] then show it is strictly increasing for x > 1
I really don't understant what you want to say with that. Can you explain.

5. Originally Posted by usual_suspect
I really don't understant what you want to say with that. Can you explain.
since $\lim_{x \to 0^+} - \ln x = \infty$ we have that $f(x) = x - \ln x - 1 > 0$ for $x \in (0,1)$

when $x = 1$ we have $f(1) = 1 - 0 - 1 = 0$

thus, $f(x) \ge 0$ for $x \in (0,1]$

now, $f'(x) = 1 - \frac 1x$. obviously $f'(x) > 0$ for all $x > 1$

thus, $f(x)$ is increasing on $(1, \infty)$. which means $f(x) \ge 0$ for $x \in (1, \infty)$

thus, for $x \in (0, \infty)$ we have that $f(x) \ge 0$

but $f(x) = x - \ln x - 1$, so this means $x - \ln x - 1 \ge 0$

adding $\ln x + 1$ to both sides, we get $x \ge \ln x + 1$ for all $x > 0$ as desired

6. Abscissa?. Why not just say x=c. Wonder why some texts insist on showing off with Rube Goldbergian jargon?.

If I am reading this correctly, it appears to be a problem related to the Mean Value Theroem.

Use

$\frac{f(b)-f(a)}{b-a}=\frac{(e-ln(e))-(1-ln(1))}{e-1} = \frac{e-2}{e-1}$

$\frac{d}{dx}[x-ln(x)]=1-\frac{1}{x}$

To find the point c, which by the MVT should be between 1 and e:

$1-\frac{1}{c}=\frac{e-2}{e-1}$

Solve for c.

Here's the graph with the tangent line and the secant line. Note they are parallel as per the MVT.

You can now find the equation of your tangent line, if needed.

7. Originally Posted by galactus
Abscissa?. Why not just say x=c. Wonder why some texts insist on showing off with Rube Goldbergian jargon?.
Here in portugal abscissa (or how we say here abcissa) is a very a term used much times. I didn't noticed that you guys don't use it. English is not my native English, so sometimes I do some mistakes.