Easy Graph question (multiple choice) (1 Viewer)

[click here]

 

[darkliight]

On the interval (-oo, -3) the graph is increasing, so f'(x) > 0 right ? This rules out B.
On the interval (-3, -1) the graph is decreasing, so f'(x) < 0 right ? This rules out C.
On the interval (-1, oo) the graph is increasing, so f'(x) > 0 again and this rules out D.

 

[click here]

Thanks mate! I got it now

Also another completely different question I am stuck on

If t is any real number, enter the minimum value of
y(t) = 1.4 + 0.7 sin(3t)

 

[darkliight]

Well -1 <= sin(3t) <= 1.

So 1.4 + 0.7 * (-1) = 0.7 would be the minimum value. Can you give a value for t when this happens?

 

[withoutaface]

t=pi/2 or -pi/6, depending on which way you're inclined.

 

[pkc]

How come its not t=-pi/12 ??

isnt sin(-pi/4) =-1 ?

 

[click here]

Awesome man! Thanks for your explaination, it cleared things up, and I can't believe how easy that was

much <3 to darklight

 

[webby234]

How come its not t=-pi/12 ??

isnt sin(-pi/4) =-1 ?

No, sin (-pi/4) = -1/rt2

sin (-pi/2) = -1

 

[pkc]

Hey youre right!
Was confusing 90deg with pi/4.Doh!
Never did like radians.

Thanks:wave:

 

[sasquatch]

Would you need to say that as the equation y(t) = 1.4 + 0.7sin(3t) has multiple minimum values, which are -π/6, π/2, 11π/6, ... so would the answer be

y(t) is a minimum for t = -π/6 ± 2π/3