This was the first problem on my test

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In summary: = \frac{f(b+\frac{1}{2}) - f(a+\frac{1}{2})}{b+\frac{1}{2}} = \frac{f(b+\frac{1}{2}) - f(a+\frac{1}{2})}{2} = \frac{f(b+\frac{1}{2})}{4} = \frac{1}{4}
  • #1
SELFMADE
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Homework Statement



Find the average rate of change of function g(t)=3+tan(t) under interval [-pi/4;pi/4]

Homework Equations





The Attempt at a Solution



I am stuck at (-tan(-pi/4))/pi
 
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  • #2
What are sin(pi/4) and cos(pi/4)?
 
  • #3
Same. So tan(pi/4) is 1?
 
  • #5
Of course, you say your difficulty is with
[tex]-tan(-\pi/4)/\pi= -\frac{sin(-\pi/4)}{cos(-\pi/4)}[/tex]
What is that?
 
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  • #6
So my next question is: what happens to the sine and the cosine when you put a minus in the argument. In other words: how does sin(-x) relate to sin(x) and how does cos(-x) relate to cos(x)?
 
  • #7
It is the inverse
 
  • #8
SELFMADE said:
It is the inverse

-x => you measure clockwise from the horizontal axis, draw for quadrants and draw an acute angle clockwise from the positive horizontal axis. What quadrant does it lie in? Is sin(x) positive or negative in this quadrant?
 
  • #9
it lies on IV quadrant. sin is only positive in quadrant II, so it is negative.
 
  • #10
SELFMADE said:
it lies on IV quadrant. sin is only positive in quadrant II, so it is negative.

thus sin(-x)=-sinx, do the same for cos(x)
 
  • #11
SELFMADE said:
it lies on IV quadrant. sin is only positive in quadrant II, so it is negative.

? Sine is positive in the I and III quadrants. It is negative in the II and IV quadrants.
 
  • #12
SELFMADE said:
it lies on IV quadrant. sin is only positive in quadrant II, so it is negative.

? Sine is positive in the I and III quadrants. It is negative in the II and IV quadrants.

But, as Rockfreak667 pointed out, it is much simpler to use the fact that sine is an odd function, cosine an even function, that to worry about "quadrants".
 
  • #13
HallsofIvy said:
? Sine is positive in the I and III quadrants. It is negative in the II and IV quadrants.

But, as Rockfreak667 pointed out, it is much simpler to use the fact that sine is an odd function, cosine an even function, that to worry about "quadrants".

That's half right. Sine is positive in the quadrants I and II, and negative in quadrants III and IV. Cosine is positive in quadrants I and IV, and negative in quadrants II and III. As a result, tangent is positive in quadrants I and III, and negative in quadrants II and IV.
 
  • #14
So the answer is -1/pi?

Got 88 on my first calculus test. If I had this problem correct would have gotten an A.
 
  • #15
SELFMADE said:
So the answer is -1/pi?
I don't think so. Your answer should be positive. The average rate of change of a function f on an interval [a, b] is
[tex]\frac{f(b) - f(a)}{b - a}[/tex]
 

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