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mparra4761
13.07.2019 •
Mathematics
Consider the function on the interval (0, 2π). f(x) = sin x + cos x (a) find the open intervals on which the function is increasing or decreasing. (enter your answers using interval notation.)
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Ответ:
Increasing in x∈(0,π/4)∪(5π/4,2π) decreasing in(π/4,5π/4)
Step-by-step explanation:
given f(x) = sin(x) + cos(x)
f(x) can be rewritten as![\sqrt{2} [\frac{sin(x)}{\sqrt{2} }+\frac{cos(x)}{\sqrt{2} } ]..................(a)\\\\\ \frac{1}{\sqrt{2} } = cos(45) = sin(45)\\\\](/tpl/images/0083/6226/7aa1c.png)
Using these result in equation a we get
f(x) =![\sqrt{2} [ cos(45)sin(x)+sin(45)cos(x)]\\\\= \sqrt{2} [sin(45+x)]..........(b)](/tpl/images/0083/6226/6aa17.png)
Now we know that for derivative with respect to dependent variable is positive for an increasing function
Differentiating b on both sides with respect to x we get
f '(x) =![f '(x)=\sqrt{2} \frac{dsin(45+x)}{dx}\\ \\f'(x)=\sqrt{2} cos(45+x)\\\\f'(x)0=\sqrt{2} cos(45+x)0](/tpl/images/0083/6226/d7192.png)
where x∈(0,2π)
we know that cox(x) > 0 for x∈[0,π/2]∪[3π/2,2π]
Thus for cos(π/4+x)>0 we should have
1) π/4 + x < π/2 => x<π/4 => x∈[0,π/4]
2) π/4 + x > 3π/2 => x > 5π/4 => x∈[5π/4,2π]
from conditions 1 and 2 we have x∈(0,π/4)∪(5π/4,2π)
Thus the function is decreasing in x∈(π/4,5π/4)
Ответ:
12 do both hurdles and long jump
Step-by-step explanation:
3/8 of the team does hurdles, so that number is 3/8·40 = 15.
Of those 15, 4/5 do long jump, so there are ...
4/5·15 = 12 . . . hurdlers who do long jump