derivatives-differentiation

W

wonky-faint

New member
Joined
Aug 28, 2006
Messages
17
this is beginning calculus.

Question: Consider the functions y=f(x)

f(x)=-x^2
f(x)=-1/x
f(x)=(x-2)^1/3
f(x)=(2-x)^1/3
f(x)=(1-x)^1/2
f(x)=(x-1)^1/4

a) graph y1 = f(x) and y2= NDERIV(y1) in the same viewing window
b) for what values of x, if any, does y1' fail to exist? why? how does the graph of y2 = NDER y1 help answer this question?
c) for what values of x, if any, is y1' positive? zero? negative?
d) for what value of x, if any, is the slope of the line tangent to the curve y1=f(x) positive? zero? negative?
e) over what intervals of x-values, if any, does the function y1=f(x) increase as x increases? decrease as x increases? how is this connected with what you found in part (c)?


can you please show me for any one of the f(x) just so i have an example to work with to see what im doing?

this was my attempt at the f(x)=-x^2:

b)y1' exists for all values. the graph of y2 =NDERy1 is continuous.
c)y1 is positive for all values of x>0.
d)slope of tangent line is negative for all values.
e) i have no clue......
 
It appears that you have a great many questions.
Please respond with a post that asks a single question that may answer several questions.
Remember, we do not give tutorials.
We do not do you homework for you.
We do try to help you with work work!
 
wonky-faint said:
can you please show me for any one of the f(x) just so i have an example to work with to see what im doing?
Click to expand...

i only asked for the answer to one question.

but if you insist:

Question: Consider the functions y=f(x)

f(x)=(x-1)^1/4

a) graph y1 = f(x) and y2= NDERIV(y1) in the same viewing window
b) for what values of x, if any, does y1' fail to exist? why? how does the graph of y2 = NDER y1 help answer this question?
c) for what values of x, if any, is y1' positive? zero? negative?
d) for what value of x, if any, is the slope of the line tangent to the curve y1=f(x) positive? zero? negative?
e) over what intervals of x-values, if any, does the function y1=f(x) increase as x increases? decrease as x increases? how is this connected with what you found in part (c)?


can anyone help me with that?
 
a) To learn how to operate your calculator, please review your owners manual. If you have lost your manual, you can download a new one from the manufacturer's web site.

b) Take the derivative. Look at the expression for the derivative. Think back on what you learned in algebra about where functions are defined, and where they are not (such as negatives inside square roots, or zeroes in denominators). Look at the picture on the calculator screen. Compare the graph with the algebraic solution you just found.

c) Return to the derivative function that you found for part (b). Think back on what you learned in algebra about "f(x) = 0" meaning the function was crossing the x-axis, "f(x) > 0" meaning it was above the axis, and "f(x) < 0" meaning it was below. Solve the equation y1' = 0, or else work from the picture on the calculator screen.

d) Since "the slope of the tangent line" is the value of the derivative, y1', at any given point, use the answers you just got for part (c).

e) Look at the graph of y1. You've found the zeroes of y1', the derivative, which divide the number line into intervals of increase or decrease for y1. Use the graph and the information from parts (c) and (d), along with what you've learned (since algebra) about the First Derivative Test.

Eliz.
 
oh now THAT helped!!! THANKS!! i couldnt figure out what to put after NDERIV y1, X,

but sorta late. i already had to turn it in.
 
You must log in or register to reply here.
Top