anonymous
  • anonymous
For what intervals is f(x) = 2x^4 – 4x^2 + 6 increasing? A. The graph is increasing on the intervals (-1, 0) and (1, ∞). B. The graph is increasing on the interval (1, ∞). C. The graph is increasing on the interval (-1, 0). D. The graph is increasing on the intervals ( -∞ , -1) and (0, 1).
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
You can differentiate the function and it is increasing if f ' (x)>0.
anonymous
  • anonymous
huh?
anonymous
  • anonymous
@electrokid can u help?

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anonymous
  • anonymous
Differentiating f(x) we get \[f'(x)=8x^3 - 8x=8x(x^2-1)\] That is positive for all x>1.
anonymous
  • anonymous
ok
anonymous
  • anonymous
so what do i do next?
anonymous
  • anonymous
That's it :)
anonymous
  • anonymous
ohh so it will be D?
anonymous
  • anonymous
No...D includes negative numbers. You need the one thats only x>1.
anonymous
  • anonymous
|dw:1364917996121:dw| get the critical points you see the four intervals? take an arbitrary number from each interval and check the signs of the derivative
anonymous
  • anonymous
Ohh so its B
anonymous
  • anonymous
Yep :)
anonymous
  • anonymous
nope
anonymous
  • anonymous
what do you mean nope?
anonymous
  • anonymous
so its wrong?
anonymous
  • anonymous
Its right....
anonymous
  • anonymous
yep. B is not the answer
anonymous
  • anonymous
check the interval (-1,0)
anonymous
  • anonymous
in math, there are no shortcuts. follow the "yellow brick road"
anonymous
  • anonymous
ok
anonymous
  • anonymous
do you see the mistake?
anonymous
  • anonymous
yea
anonymous
  • anonymous
so, what is the answer?
anonymous
  • anonymous
i got C
anonymous
  • anonymous
follow the steps. did you use the number line that I drew?
anonymous
  • anonymous
yea hold on
anonymous
  • anonymous
so the critical points are -1, 0 and 1
anonymous
  • anonymous
okay
anonymous
  • anonymous
sorry. yes. you are correct. "3" points, -1, 0 and 1
anonymous
  • anonymous
lol ok
anonymous
  • anonymous
|dw:1364919097503:dw|
anonymous
  • anonymous
interval (-inf, -1) plug in x=-2 and find f'(x)
anonymous
  • anonymous
plug in -2 into the problem?
anonymous
  • anonymous
yep
anonymous
  • anonymous
into (x - 2)^2?
anonymous
  • anonymous
into the entire thing f'(x)
anonymous
  • anonymous
ok
anonymous
  • anonymous
idk i got 16?
anonymous
  • anonymous
and the derivative i got 0
anonymous
  • anonymous
\[f'(-2)=8(-2)[(-2)^2-1]=?\]
anonymous
  • anonymous
ohh i got -48
anonymous
  • anonymous
good.|dw:1364919435696:dw| now, we check interval (-1,0) find \[f'(-0.5)\]
anonymous
  • anonymous
okay i get 0
anonymous
  • anonymous
nope try again.
anonymous
  • anonymous
you mean find the derivative of -0.5 ?
anonymous
  • anonymous
derivative "at" x=-0.5
anonymous
  • anonymous
ohhh ok
anonymous
  • anonymous
yep. what did you get?
anonymous
  • anonymous
i'm still getting 0 :/
anonymous
  • anonymous
\[f'(-0.5)=8(-0.5)[(-0.5)^2-1]=?\]
anonymous
  • anonymous
ok i get 3
anonymous
  • anonymous
good.|dw:1364919843444:dw| now we check interval (0,1) lets take x = 0.5 similarly, find \[f'(0.5)\]
anonymous
  • anonymous
okay
anonymous
  • anonymous
i get -3
anonymous
  • anonymous
great |dw:1364919912982:dw| now the final interval (1,+infty) we take x= 2 \[f'(2)=?\]
anonymous
  • anonymous
ok i get 48
anonymous
  • anonymous
perfect. |dw:1364919991818:dw| now, f(x) is increasing when f'(x) >0 (i.e., f'(x) is positive) and f(x) is decreasing when f'(x) < 0 (i.e., f'(x) is negative)
anonymous
  • anonymous
so, in what intervals do you see f'(x) positive?
anonymous
  • anonymous
ok in 0,1 and 1, + infi
anonymous
  • anonymous
great. which option says that?
anonymous
  • anonymous
A!
anonymous
  • anonymous
and that is your answer yay
anonymous
  • anonymous
thanks :)
anonymous
  • anonymous
follow the same procedure for the other one,
anonymous
  • anonymous
ok

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