local min/max problem
- Thread starter packerman
- Start date
f'(x) = 3x^2 - 2bx + 5 = 0
^Keep the B right??
Well I did manage to get this far when I attempted this problem but I have no idea what to do next or if I even got the differential right. Will the variable 'b' be equal to 0 when you differentiate thus negating -2bx or will it be treated as a constant?
Any additional help would be appreciated!
^Keep the B right??
Well I did manage to get this far when I attempted this problem but I have no idea what to do next or if I even got the differential right. Will the variable 'b' be equal to 0 when you differentiate thus negating -2bx or will it be treated as a constant?
Any additional help would be appreciated!
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- Apr 12, 2005
- Messages
- 11,339
You're just not paying attention. Why did you write "=0" and where is the "/6"?
One zero is:
\(\displaystyle \L\,x\,=\,\frac{-(-2b)+\sqrt{(-2b)^{2}-4*3*5}}{2*3}\,=\,\frac{b+\sqrt{b^{2}-15}}{3}\)
What is that thing you wrote? It does not indicate a knowledge of the Quadratic Formula.
Very Rough Overview
1) Given f(x)
2) Find f'(x)
3) Find x=x<sub>i</sub> such that f'(x<sub>i</sub>) = 0
4) x=x<sub>i</sub> are the x-coordiantes of local minima and maxima.
5) The minima and maxima are (x<sub>i</sub>,f(x<sub>i</sub>))
If this problem were a snake, it would have bitten you several times.
Are you SURE you're ready for calculus?
One zero is:
\(\displaystyle \L\,x\,=\,\frac{-(-2b)+\sqrt{(-2b)^{2}-4*3*5}}{2*3}\,=\,\frac{b+\sqrt{b^{2}-15}}{3}\)
What is that thing you wrote? It does not indicate a knowledge of the Quadratic Formula.
Very Rough Overview
1) Given f(x)
2) Find f'(x)
3) Find x=x<sub>i</sub> such that f'(x<sub>i</sub>) = 0
4) x=x<sub>i</sub> are the x-coordiantes of local minima and maxima.
5) The minima and maxima are (x<sub>i</sub>,f(x<sub>i</sub>))
If this problem were a snake, it would have bitten you several times.
Are you SURE you're ready for calculus?