More Calculus!?

Question

the curve in the figure represents the graph of f, where f(x)=(x^2)-2x, for all real numbers x

a.On the axes provided. sketch the graph of y= /f(x)/ (absolute value)


b. Determine whether the derivative /f(x)/ exists at x=0. Justify your answer

c. on the axes provided sketch the graph of y=f(/x/)

d. Determine whether the derivative f(/x/) is continuous at x=0. Justify your answer

Answer 1

(a) First, draw the graph of f(x) = x^2 - 2x.
This should be a parabola with x-intercepts x = 0 and x = 2. The vertex should be located at (1,-1).

Then, whenever the graph falls under the x-axis, reflect that portion of the graph along the x-axis. Whenever the graph is above the axis, don't modify.

Answer 2

a) you need to draw the curve y=/x^2 - 2x/:
you can work out the points e.g.
x = 1 y = /1-2/=1
x = 2 y = /4-4/ = 0 etc. (remembering to take the positive values of the r.h.s of the equation

b) find dy/dx where y = x^2 - 2x
put x = 0 into the equation
if theres a value then the curve has a gradient of this value at that point

c) draw the graph of y = x^2-2x for positiv values of x

d) hmm... similar idea to part b)

Answer 3

x=-2,y=8
x=-1,y=3
x=0,y=0
x=1,y=0
x=2,y=0
x=3,y=3
x=4,y=8
a]graph can be sketched using above values
b]no
c]graph is same as above
d]F'[1]=0
F'[0]=-2
F'[-1]=0
not continuous