Verified answer

a) It is a single function, so it is "the same curve". The right half of the graph will be a straight line from the point (0,0) to (2,2), another straight line from the point (2,2), to (4,0). Since the function is defined to be an "even" function, the left half of the graph will be a mirror image of the right: a straight line from the point (0,0) to (-2,2) and another straight line from the point (-2, 2) to (-4,0). The complete graph will look like two teeth of a sawtooth wave.

b) You can find these values graphically by drawing the line y = (x+4)/4 on your graph and noting where the line intersects the graph of the function. You can also do it analytically as follows:

x = (x+4)/4 => 4x = x + 4 => 3x = 4 => x = 4/3 is in the interval 0 ≤ x < 2 so one value of x which meets the requirement is x = 4/3

4-x = (x+4)/4 => 16-4x = x+4 => -5x = -12 => x = 12/5 is in the interval 2 ≤ x ≤ 4 so another value of x which meets the requirement is x = 12/5

And since it is an even function, f(x) = f(-x):

-x = (x+4)/4 = -4x = x+4 => x = -4/5 is in the interval -2< x ≤ 0 so it meets the requirement

x-4 = (x+4)/4 => 4x-16 = x+4 => 3x = -12 => x = -4 is in the interval -4 ≤ x ≤ 2 so it meets the requirement.

The 4 values of x which meet the requirement that f(x) = (x+4)/4 are x = -4, -4/5, 4/3, 12/5.

...

Yes, you have to draw it on the same graph, since it is one function

a)

It's a fairly simple graph to draw, since all "pieces" of the function are linear. So you just have to find the two endpoints of each interval, and draw a straight line between them.

f(x) = x . . . . . . . 0 ≤ x < 2 . . . . . . . endpoints (0,0)-closed and (2,2)-open

f(x) = 4 - x . . . . 2 ≤ x ≤ 4 . . . . . . . endpoints (2,2)-closed and (4,0)-closed

Now when you have an open endpoint such as (2,2) on interval [0, 2), you usually draw this with an open circle (non filled in circle). But since (2,2) is then a closed endpoint on interval [2,4], you must then draw this point with a filled-in circle.

So you get a graph like this, but using only triangular part below point of intersection, and including point of intersection

http://www.wolframalpha.com/input/?i=f%28x%29+%3D+...

Now to graph from -4 to 0, just graph mirror image of this.

Graph should look like this (except a little flatter) with top of graph at y = 2

. . . . . . . . y-axis

. . . . . . . . . . |₂

. . . . . . /\. . . | . . /\

. . . . . / .. \. . |₁. / .. \

. . . . / ...... \. | ./ ...... \

___/_____\|/_____\_______ x-axis

. . -4 . -2 . . 0 . . 2 . . 4

b)

Piecewise function f(x):

f(x) = x+4 . . . . . . . -4 ≤ x ≤ -2

f(x) = -x . . . . . . . . -2 < x < 0

f(x) = x. . . . . . . . . . 0 ≤ x < 2

f(x) = 4 - x . . . . . . . 2 ≤ x ≤ 4

So we solve f(x) = (x+4)/4 for each "piece" of function

On interval -4 ≤ x ≤ -2

x + 4 = (x + 4) / 4

4x + 16 = x + 4

3x = -12

x = -4

On interval -2 < x < 0

-x = (x + 4) / 4

-4x = x + 4

-5x = 4

x = -4/5

On interval 0 ≤ x < 2

x = (x + 4) / 4

4x = x + 4

3x = 4

x = 4/3

On interval 2 ≤ x ≤ 4

4 - x = (x + 4) / 4

16 - 4x = x + 4

-5x = -12

x = 12/5

All values of x that we found are within their respective intervals, so they are all valid

x = -4, -4/5, 4/3, 12/5

Wow, it rather is been an prolonged time for me on those, elementary algebra subject concerns, yet would not the "style" in fact be {-27,21} which isn't listed in the outcomes? otherwise, if i'm incorrect, which I likely am, I settle for as real with different posters and answer "C". **yet in a various thank you to think of of roughly those is graphically, that could help placed you previous to the class. subsequently you're able to have a line with the slope of -8x then in fact pass it up +5 and you gets an concept what for sure like. From there you may medical care at as rapidly as for any fee. in diverse words, i ought to "rewrite" the equation in my head and image of it as y= -8x+5 in some circumstances it enables!