can someone steer me in the right direction on how to solve these? thanks

for the first one.. start by plotting the points they give u.

then use the limit they gave u to sketch out the graph.
limit as f(x) approachs infinity equals zero..

and for #6 u should know what a e^x graph should look like by now. that -1 is just a shift in the vertical direction.. and u should also know what sinx graph by now. just plot it lol

This post was edited by FamilyGuyViewer on Feb 15 2014 12:38am

for first one is f'1 implying the derivitive? so just plot 2,1 -2,0 etc?

the very basics of it would be to plot the given points (2,1), (-2,0), then simply draw straight lines to connect them

f' is commonly used to denote the first derivative of f, f'' would be second derivative

next part would be to satisfy all the given conditions by redrawing the straight lines as needed

This post was edited by saber_x3 on Feb 15 2014 01:09am

are the derivitives plotted the same way? as the first 2 points? sorry my professor doesnt do a good job explaining

if you're thinking of plotting f'(-1)=0 as point (-1,0), then you have a major lack of understanding of the current material

derivative basically means slope or change.
f'(-1) =0 means the derivative at x=-1 is 0. means the slope is 0

i'd suggest you to look over notes/book, or google khan academy-> intro to calculus or so

This post was edited by saber_x3 on Feb 15 2014 02:47am

for a, a function with 2 local min/maxes would do nicely, so lets try:
f(x) = Ax^3 + Bx^2 + Cx + D

then just plug in what we know:

f(2) = A(2)^3 + B(2)^2 + C(2) + D = 1
f(2) = 8A + 4B + 2C + D = 1

f(-2) = A(-2)^3 + B(-2)^2 + C(-2) + D = 0
f(-2) = -8A + 4B -2C + D = 0

f'(x) = 3Ax^2 + 2Bx + C

f'(-1) = 3A(-1)^2 + 2B(-1) + C = 0
f'(-1) = 3A - 2B + C = 0

f'(2) = 3A(2)^2 + 2B(2) + C = 0
f'(2) = 12A + 4B + C = 0


4 equations, 4 unknowns

gives:

A = (-1/8)
B = (3/16)
C = (3/4)
D = (-1/4)

f(x) = (-1/8)x^3 + (3/16)x^2 + (3/4)x + (-1/4)

seems to meet all the requirements.

b is wrong, you were fine up to and including cancelling the (x+2) terms, at that point you should just substitute x=-2 and evaluate the expression, I'm not sure what you actually did after this...

c is really easy, just plug in pi and evaluate it

same with d

e you made a mistake in the last step (when rationalizing your denominator)

on 7 you wrote down the correct formula, but then didn't actually use it.... Follow the formula...

6, well they are both pieces of the piecewise function are continuous, so the only possible problem would be where they are joined together, so establish they meet at this location and you are in business

^^^

oh and for the first 2, I was over-thinking it, you don't need to come up with the function, you just need to fudge up a graph that meets those requirements! Just draw that sucker! For (a) just remember when you are at a local min/max, the derivite is equal to 0, so just make sure you hit the 2 points listed, and you have a "slope" of 0 when you pass though the areas where the derivative is zero. I feel dumb