# Thread: [SOLVED] Finding Extreme Values...

1. ## [SOLVED] Finding Extreme Values...

i know i am supposed to find the derivative of an equation to find [somehow] the extreme points.

but how? for example:
$f(x)= \frac {1}{x} + \ln(x)$

i would get this right?
$\frac {-1}{x^2} + \frac {1}{x}$
they give me an interval as well... $0.5 \leq x \leq 4$

what do i do?

2. Hello,
Originally Posted by >_<SHY_GUY>_<
i know i am supposed to find the derivative of an equation to find [somehow] the extreme points.

but how? for example:
$f(x)= \frac {1}{x} + \ln(x)$

i would get this right?
$\frac {-1}{x^2} + \frac {1}{x}$
they give me an interval as well... $0.5 \leq x \leq 4$

what do i do?

Now you have to solve for $\frac{-1}{x^2}+\frac 1x=0$
Multiply each side by $x^2$ (which is possible since $x\neq 0$) and you'll get $-1+x=0$, which is easier to deal with

The same trick can be used if you're looking for the values for which it's >0.

Spoiler:

You will see that $f'(x)>0 \Leftrightarrow x>1$
This means that on (1,4), it's increasing.
And on (.5,1), it's decreasing.
Thus there is a minimum at x=1 (draw a rough sketch)

3. Originally Posted by Moo
Hello,

Now you have to solve for $\frac{-1}{x^2}+\frac 1x=0$
Multiply each side by $x^2$ (which is possible since $x\neq 0$) and you'll get $-1+x=0$, which is easier to deal with

The same trick can be used if you're looking for the values for which it's >0.

Spoiler:

You will see that $f'(x)>0 \Leftrightarrow x>1$
This means that on (1,4), it's increasing.
And on (.5,1), it's decreasing.
Thus there is a minimum at x=1 (draw a rough sketch)
Hey ,
ok, i get most of it, except the how you got 4 to get the max value at (1,4)

there is also a local maximum too, and thats where i get lost to, can we figure out both local max and min from the derivative?

Merci Beaucoup

4. Originally Posted by >_<SHY_GUY>_<
i know i am supposed to find the derivative of an equation to find [somehow] the extreme points.

but how? for example:
$f(x)= \frac {1}{x} + \ln(x)$

i would get this right?
$\frac {-1}{x^2} + \frac {1}{x}$
they give me an interval as well... $0.5 \leq x \leq 4$

what do i do?
set f'(x) = 0

$\frac {-1}{x^2} + \frac {1}{x} = 0$

$x = 1$ is a critical value

evaluate the second derivative at x = 1 ...

$f''(x) = \frac{2}{x^3} - \frac{1}{x^2}
$

$f''(1) = 2 - 1 = 1 > 0$

the graph of f is concave up ... therefore, f has a minimum at x = 1

note that f also has endpoint extrema (maximums) at x = 0.5 and x = 4

5. Originally Posted by skeeter
set f'(x) = 0

$\frac {-1}{x^2} + \frac {1}{x} = 0$

$x = 1$ is a critical value

evaluate the second derivative at x = 1 ...

$f''(x) = \frac{2}{x^3} - \frac{1}{x^2}
$

$f''(1) = 2 - 1 = 1 > 0$

the graph of f is concave up ... therefore, f has a minimum at x = 1

note that f also has endpoint extrema (maximums) at x = 0.5 and x = 4
that was another great concept i tried to understand: concavity.
and it was really nice of you to point that out.

i my seem a bit, naive , but that is one thing i don't get,
how to find the critical points of a graph. here it's 1, but why?

6. Originally Posted by >_<SHY_GUY>_<
how to find the critical points of a graph. here it's 1, but why?
$-\frac{1}{x^2} + \frac{1}{x} = 0$

since $x \ne 0$ , multiply every term by $x^2$ ...

$-1 + x = 0$

$x = 1$

7. Originally Posted by skeeter
$-\frac{1}{x^2} + \frac{1}{x} = 0$

since $x \ne 0$ , multiply every term by $x^2$ ...

$-1 + x = 0$

$x = 1$
is there any exact way of telling and figurring out the local extremas?

8. Originally Posted by >_<SHY_GUY>_<
is there any exact way of telling and figurring out the local extremas?
The points at the boundaries of the interval are candidates to be (local) extremas

Provided that your interval is bounded, though.

9. Originally Posted by Moo
The points at the boundaries of the interval are candidates to be (local) extremas

Provided that your interval is bounded, though.
oh!
i understand it much better...

ok, lets say i have
$\ln (x+1) , 0\leq x \leq 3$
the derivative would give me:
$\frac {x}{x+1}$
make itequal to zero, multiply both sides by x+1, giving you $x = 0$
meaning the min. is at X=0, and that the max. is at X=3

Right?

10. Originally Posted by >_<SHY_GUY>_<
oh!
i understand it much better...

ok, lets say i have
$\ln (x+1) , 0\leq x \leq 3$
the derivative would give me:
$\frac {x}{x+1}$
make itequal to zero, multiply both sides by x+1, giving you $x = 0$
meaning the min. is at X=0, and that the max. is at X=3

Right?
no.

the derivative of $y = \ln(x+1)$ is $y' = \frac{1}{x+1}$

$\frac{1}{x+1} \ne 0$ for any x in the domain of y

11. Hmm the derivative is $\frac{1}{x+1}$

Now you can see that over [0,3], it's always positive. So f is always increasing.
Meaning, as you said , that the min is at x=0 and the max is at x=3.

Be careful, it's not always the case !! This works under given conditions only !
The best way to check is probably to draw a graph

12. Originally Posted by Moo
Hmm the derivative is $\frac{1}{x+1}$

Now you can see that over [0,3], it's always positive. So f is always increasing.
Meaning, as you said , that the min is at x=0 and the max is at x=3.

Be careful, it's not always the case !! This works under given conditions only !
The best way to check is probably to draw a graph
Damn, i saw my mistake
thank you though

umm, there is this one method my teacher vaguely explain, something about a number line, saying that if it is greater than something, it increases, meaning it is a max. point, and if it decreases, it is a min point.

do you know this?

13. You know that $a\leq x\leq b$
And if the function f is increasing, then if $c\leq d$, thenn $f(c)\leq f(d)$
If it's decreasing, then $f(c)\geq f(d)$

This is why you sometimes have to consider the boundary points.

And this is the principle of studying the sign of the derivative (think about it ! ^^)

14. Originally Posted by Moo
You know that $a\leq x\leq b$
And if the function f is increasing, then if $c\leq d$, thenn $f(c)\leq f(d)$
If it's decreasing, then $f(c)\geq f(d)$

This is why you sometimes have to consider the boundary points.

And this is the principle of studying the sign of the derivative (think about it ! ^^)
hmmmm, i get it a bit, but i see why you made a point there.

ummm , i know it seems i ask too much :/

but this chapter confuses me alot,
especially when they throw me off with trig functions and especially exponents.

im currently stuck with this one:

$f(x) = x^{2/5} , -3 \leq x \leq 1$

all ive done so far is

$\frac {2}{3} x^{-3/5}$

but i dont know what to do next

15. Nope, the derivative is $\frac{2}{\color{red}5} \cdot x^{-3/5}$

Note that if x=0, the derivative is not defined. Though the function exists.

Since the derivative would be infinity, this means that there is a vertical asymptote to the curve of f. Sketch it on your calculator

Then you can see that if you consider (0,infinity) (0 not included), $f'(x)>0$. So f is strictly increasing over [0,infinity) (0 is included - because f is defined at x=0), meaning that 0 is a minimum.
The maximum is $\lim_{x\to\infty} f(x)=\infty$. so there is no maximum.

For (-infinity,0), you can just note that $f(-x)=f(x)$, because $x^{2/5}=(x^{1/5})^2=(x^2)^{1/5}=(-x)^{2/5}$
So 0 is also a minimum value for f over (-infinity,0]

I really hope this helps you grasping the problems... I'm not that good at explaining these things

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