1. ## solving equations

I was looking for critical points and found my f'(x)=-12/(4x^2+1)(\sqrt{4x^2+1})

Now I want to solve that equation , by setting
-12/(4x^2+1)(\sqrt{4x^2+1}=0

will there be any solution for x? or this is an equation that has no solution?

2. ## Re: solving equations

Originally Posted by ameerulislam
I was looking for critical points and found my f'(x)=-12/(4x^2+1)(\sqrt{4x^2+1})

Now I want to solve that equation , by setting
-12/(4x^2+1)(\sqrt{4x^2+1}=0

will there be any solution for x? or this is an equation that has no solution?
Is it that equation that you meant to write?

$\displaystyle f'(x)= \frac{-12\sqrt{4x^2+1}}{4x^2+1}$

3. ## Re: solving equations

Originally Posted by dokrbb
Is it that equation that you meant to write?

$\displaystyle f'(x)= \frac{-12\sqrt{4x^2+1}}{4x^2+1}$
No, only -12 is up there.

$\displaystyle f'(x)= \frac{-12}{(\sqrt{4x^2+1})(4x^2+1)}$

4. ## Re: solving equations

so

what would be the solution for

$\displaystyle \frac{-12}{(\sqrt{4x^2+1})(4x^2+1)} =0$

5. ## Re: solving equations

Originally Posted by ameerulislam
so

what would be the solution for

$\displaystyle \frac{-12}{(\sqrt{4x^2+1})(4x^2+1)} =0$
I'm not sure, but working with the denominator ...

$\displaystyle \frac{-12}{({4x^2+1}^{\frac{1}{2}})(4x^2+1)^{1}}=0$

it gives you this $\displaystyle \frac{-12}{{4x^2+1}^{\frac{3}{2}}}=0$ you have a constant at the numerator, and apparently the denominator is never $\displaystyle 0$ because of
$\displaystyle {4x^2+1}^{\frac{3}{2}}}$

could you provide as with the original $\displaystyle f(x)$ ?

dokrbb

6. ## Re: solving equations

sure, this was the question

f(x)= 3/sqrt(4x^2 + 1) ; [-1,1] , the question asked to find the absolute max and min.

7. ## Re: solving equations

you were supposed to get this as derivative $\displaystyle \frac{-12x}{{4(x)^2+1}^{\frac{3}{2}}}$

and because of $\displaystyle -12x$ the only critical point would be at $\displaystyle x=0$

can you solve for x now

8. ## Re: solving equations

Originally Posted by dokrbb
you were supposed to get this as derivative $\displaystyle \frac{-12x}{{4(x)^2+1}^{\frac{3}{2}}}$

and because of $\displaystyle -12x$ the only critical point would be at $\displaystyle x=0$

can you solve for x now
oh.. you are right...

Ok now we have 3 candidates for relative maxima

one CP: 0,
and 2 End points: 1 and -1

if I plug those numbers on the original equation

$\displaystyle \frac{3}{\sqrt{(4x^2 + 1)}}$

i get

For 0 , I get 3
for -1, I get $\displaystyle \frac{3}{\sqrt{5}}$

for 1, I get $\displaystyle \frac{3}{\sqrt{5}}$

Am I right? so absolute maximum is 0 and absolute minimum is -1 and 1 ?

But in my book (in the answer section) it says something else
maximum value $\displaystyle \frac{3}{\sqrt{5}}$ at x=1

minimum value $\displaystyle -\frac{3}{\sqrt{5}}$ at x=-1

I have no idea how did they get $\displaystyle -\frac{3}{\sqrt{5}}$ , that x was squared right ?

9. ## Re: solving equations

Originally Posted by ameerulislam
oh.. you are right...

Ok now we have 3 candidates for relative maxima

one CP: 0,
and 2 End points: 1 and -1

if I plug those numbers on the original equation

$\displaystyle \frac{3}{\sqrt{(4x^2 + 1)}}$

i get

For 0 , I get 3
for -1, I get $\displaystyle \frac{3}{\sqrt{5}}$

for 1, I get $\displaystyle \frac{3}{\sqrt{5}}$

Am I right? so absolute maximum is 0 and absolute minimum is -1 and 1 ?

But in my book (in the answer section) it says something else
maximum value $\displaystyle \frac{3}{\sqrt{5}}$ at x=1

minimum value $\displaystyle -\frac{3}{\sqrt{5}}$ at x=-1

I have no idea how did they get $\displaystyle -\frac{3}{\sqrt{5}}$ , that x was squared right ?
I'm sure someone would correct me on this point but since your function is defined in the interval [-1, 1] the value of x=0 doesn't respect the conditions for the absolute maxima and minima, and in this case you woul like to look for maxima and minima at the endpoints.

Here is a more or less detailed description - "Restricted domains: There may be maxima and minima for a function whose domain does not include all real numbers. A real-valued function, whose domain is any set, can have a global maximum and minimum. There may also be local maxima and local minima points, but only at points of the domain set where the concept of neighborhood is defined. A neighborhood plays the role of the set of x such that |x − x∗| < ε."

dokrbb

10. ## Re: solving equations

I just used and online math calculator and I feel my observation suppose to be correct

11. ## Re: solving equations

Originally Posted by ameerulislam
I just used and online math calculator and I feel my observation suppose to be correct

Yes, but this is not for the function which is restricted on [-1, 1] interval - that's what makes the difference in your case,

see that as an example Maxima and minima - Wikipedia, the free encyclopedia

dokrbb

12. ## Re: solving equations

Your argument seems to be correct, and the graph makes it pretty obvious. There's a maximum of 3 at x=0 and minima of $\displaystyle \frac{3}{\sqrt{5}}$ at x=1 and x=-1. Since the function is restricted to $\displaystyle -1 \le x \le 1$, you ignore the parts of the graph to the right of x=1 and to the left of x=-1.

It would seem your book is wrong. Assuming you've relayed the question and solution correctly........

- Hollywood

13. ## Re: solving equations

Sorry guys.. I made another mistake here... Just got crazy before the exam :/...

The original equation had an extra x at the top..

f(x)= 3x/sqrt(4x^2 + 1); [-1,1]

grrrrr...

14. ## Re: solving equations

Originally Posted by ameerulislam
Sorry guys.. I made another mistake here... Just got crazy before the exam :/...

The original equation had an extra x at the top..

f(x)= 3x/sqrt(4x^2 + 1); [-1,1]

grrrrr...

as you see, your function is continuous at 0, and there is actually no maxima or minima, but since you have the restricted domain - there it is,

Next time just post the whole problem, not just the parts you may think a re needed/important/relevant, would be easier for you and for all another that are willing to understand the problems here,

dokrbb

15. ## Re: solving equations

Originally Posted by dokrbb