Yoshi wrote:
Can anyone help me with some calculus?
Find the average rate of change of the function over each interval.
5. f(x) = cot x, [pi/6, pi/2]
I use the definition of the slope of the tanget line thing. Slope of the secant line, whatever that is.
(f(t+h) - f(t)) / h
Problem is, when I replace cot x with 1/tan x (cause that's all I know about cotangent, that it's the reciprocal of tangent), plug everything into the equation and into my calculator (radian mode), I get an error. Answer key says there's an answer.
Being a grade 12 math student in conjunction with being a calculus student, I haven't learned this radian mode stuff yet, and I'm guessing there's some specific manual un-electronic way to solve this #5 thing up there. If anyone can roughly explain this all to me in layman's terms and help me come up with the answer in the right manner other than copying the answer key, it'd be muchly appreciated.

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[edit] One more question...this one not as weird, but still. *scratches head*
Find the slope of the curve at the indicated point.
f(x) = |x|, at x=2 and x=-3
I got so far as this:
lim
h->0 (|x+h| - |x|) / h
Remember, tan(x) = sin(x)/cos(x).
Consequently, cot(x) = 1/tan(x) = cos(x)/sin(x).
Average rate of change = delta y / delta x = [cot(pi/2) - cot(pi/6)] / (pi/2 - pi/6) = [cos(pi/2)/sin(pi/2) - cos(pi/6)/sin(pi/6)] / (pi/2 - pi/6) = (0/1 - (sqrt(3)/2)/(1/2)) / (pi / 3) =
-3*sqrt(3)/pi
f(x) = |x| can be defined by a piecewise function:
f(x) = x for all x >= 0
f(x) = -x for all x < 0
Thus, at x = 2, y = |x| has the same rate of change (derivative) as y = x, which has a slope of 1.
At x = -3, y = |x| has the same derivative as y = -x, that is, -1.
It is also probably important to note that f(x) = |x| is
not differentiable at x = 0.
Edit: If you're just starting calculus, it might be better to calculate the derivative of |x| the rigorous way...i.e. with the limit approach you began.
The idea is actually the same, but applying it works like this:
f(x) = |x|
For all positive x, i.e. all x > 0, |x| = x (you can ignore the absolute value signs because x is positive anyways). Similarly, you can ignore the absolute value signs on |x+h| because as h->0, the expression is already positive so long as x is positive. So remove the abs. value signs from the limit:
lim h->0 (|x+h| - |x|)/h = lim h->0 (x+h-x)/h = lim h->0 h/h = 1 for all x > 0.
What about for negative x, i.e. all x < 0? Well in that case |x| = -x. Similarly, |x+h| = -(x+h) if x is negative. The fact that you're "adding" h (which could just as well be negative, because h->0 from both sides) is a non-factor---the fact that it goes to zero prevents it from changing the inside of the absolute value. Replace |x| with -x and |x+h| with -(x+h):
lim h->0 (|x+h| - |x|)/h = lim h->0 (-(x+h)-(-x))/h = lim h->0 -h/h = -1 for all x < 0.
And if x = 0? Well now we have a problem. Before, we knew what to replace the absolute values with so long as x > 0 or x < 0, because we knew the value of h would not affect the sign of the expression inside of the absolute value.
However, for x = 0, h becomes important. If h approaches zero from the left (written h->0-), THEN |x+h| is negative, and therefore equals -(x+h). If h approaches zero from the right (written h->0+), THEN |x+h| = x+h.
As you can see, lim h->0- =/= lim h->0+, and therefore, the limit does not exist, and consequently, the function is not differentiable at x = 0.
Let me know if this doesn't clear things up.