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?
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?
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
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
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
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
I would show your graph now - click here 3x/sqrt(4x^2 + 1) - Wolfram|Alpha
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