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Mathematics 28 Online
OpenStudy (anonymous):

Let f(x)=6x^2−8x^4. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima).

OpenStudy (freckles):

you get to choose to do this the algebra way or the cal way?

OpenStudy (freckles):

calculus is what you are studying so let's do it that way

OpenStudy (freckles):

find f' solve f'=0 then determine where f'>0 or <0

OpenStudy (anonymous):

ok f' is 12x -32x^3 and the zeros are 0 and +/- .6123724357

OpenStudy (freckles):

oh i'm sorry i thought I seen a quadratic

OpenStudy (freckles):

I don't have my classes on didn't see the x^4 :p

OpenStudy (anonymous):

lol it's ok

OpenStudy (freckles):

but yeah correct derivative

OpenStudy (freckles):

look me check f' 's zeros

OpenStudy (freckles):

are you allowed to use approximations?

OpenStudy (freckles):

if so I don't see anything wrong with your zeros for f'

OpenStudy (anonymous):

i'm not sure but that number's also equal to +/- (sqrt(3/2)/2)

OpenStudy (freckles):

anyways just like the last one test your intervals surrounding your critical numbers

OpenStudy (anonymous):

ok so the intervals i have to plug in are: (-inf, 0) (0, .6123724357), (.6123724357, inf) Are there any more? is that even right?

OpenStudy (freckles):

\[\text{ test the following that is } \\ (-\infty, -\sqrt{\frac{3}{8}} \approx -0.6124) \\ (-\sqrt{\frac{3}{8}} \approx -0.6124,0) \\ (0, \sqrt{\frac{3}{8}} \approx 0.6124) \\ (\sqrt{\frac{3}{8}} \approx 0.6124,\infty)\] if you have n critical numbers you have (n+1) intervals to check

OpenStudy (freckles):

It isn't that bad you just have to either check the first two intervals or the last two intervals and use the fact that f' is odd for the other intervals or just enter a number from each interval

OpenStudy (freckles):

for example I chose -5 -1/2 1/2 5 as my test subjects

OpenStudy (anonymous):

ok i'm trying it

OpenStudy (michele_laino):

Hint: if we set the condition: \[f'\left( x \right) \geqslant 0\]

OpenStudy (michele_laino):

then we are led to solve these subsequent systems: \[\begin{gathered} \left\{ \begin{gathered} x \geqslant 0 \hfill \\ 3 - 8{x^2} \geqslant 0 \hfill \\ \end{gathered} \right. \hfill \\ \left\{ \begin{gathered} x \leqslant 0 \hfill \\ 3 - 8{x^2} \leqslant 0 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \]

OpenStudy (anonymous):

so f is increasing on (-inf, -.6124), (0, .6124) and decreasing on (-.6124, 0), (.6124,inf)

OpenStudy (anonymous):

do you have to always draw it out to find the relative max and min?

OpenStudy (freckles):

not really if it switches from increasing to decreasing at a critical number then that critical number is where the max occurs if it switches from decreasing to increasing at a critical number then that critical number is where the min occurs

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