Question

Need help on Functions! Medal and Fans given

Answer 1

Are you able to graph h(x) based on the piecewise definition given?

Answer 2

How do i sketch for values of x above 5 ? and what does h(x) = h(x+6) for all values of x mean ?
Because for values of x between -1 to 5, it does not equate to (x+6), so im confused.

Answer 3

@jim_thompson5910

Answer 4

\[h(x)=h(x+6)\] means it is periodic with period 6

Answer 5

the length of the interval \([1,5]\) is \(6\) so it repeats itself after that

Answer 6

h(x) = h(x+6) means that the function repeats itself every 6 units (as satellite73 points out)

so for example, if x = 0, then

h(x) = h(x+6)

h(0) = h(0+6)

h(0) = h(6)

---------

If x = 1, then

h(x) = h(x+6)

h(1) = h(1+6)

h(1) = h(7)

and so on

Answer 7

oops \([-1,5]\)

Answer 8

what that means is you simply take ALL of h(x) that is graphed from the piecewise definition and you make a complete copy and shift it 6 units to the right to get the next period of it

Answer 9

does it means that h(7) is the same as h(1) ?

Answer 10

So here is h(x) originally just based off the piecewise definition (ignore the fact that h(x) = h(x+6) for now)

Answer 11

i manage to get that too.. but only to 5

Answer 12

make a copy of that graph and shift it 6 units to the right (to get the blue piece)

Answer 13

and yes h(7) is the same as h(1)

the heights of the curves are the same at those x values

Answer 14

oh i get it now

Answer 15

so the important point is the piecewise info

Answer 16

the h(x) = h(x+6) is jus there to guide us for values higher than 5

Answer 17

sir, can i ask what if h(x) = h(x-6) ?

Answer 18

correct, it repeats forever but we only care about the first 2 pieces because -1 <= x <= 11

Answer 19

oh forgot to put on it h(x) = h(x+6), but you get the idea

Answer 20

cuz from what i learn if it is (x+6) should be shift towards negative side of x-axis

Answer 21

that is true

Answer 22

but what this is saying is that the y coordinates of the points on the curve are the same

at x = 1, the y coord is h(1)
at x = 7, the y coord is h(7)

since h(1) = h(7), the y coords are the same

apply this to EVERY x value in the domain of h(x) and you get this periodic pattern

Answer 23

oh i get it

Answer 24

thank you so much for your help jim :)

Answer 25

that wraps up part a), do you see how to do part b?

Answer 26

uhm.. part b too difficult =.=

Answer 27

I attached the graph labeled "3.png" btw

Answer 28

what do integrals represent in terms of the graph?

Answer 29

area under the graph

Answer 30

but i dont know how to form equation from the graph to do integration

Answer 31

true, that part is tricky

Answer 32

however, let's first notice that 63/2 = 31.5

so we want the area under the curve to be 31.5 square units

the area from x = 0 to x = a where a > 0

Answer 33

what is the area of 1 single trapezoid (of one period)?

Answer 34

area of trapezium is 1/2 * a * (b+c)

Answer 35

a = y-axis, b and c is x-axis

Answer 36

so what values do you plug in? the graph may help you determine this

Answer 37

area of trapezium = 1/2 * 2 * (6+2)
= 8

Answer 38

good, that is 1 trapezoid

Answer 39

we want a total area of 31.5 but we only have 8 so far

Answer 40

3 trapezoids or trapeziums give 8*3 = 24 square units

4 of them give 8*4 = 32 square units

so we need 3 whole trapezoids and then some left over

Answer 41

making sense so far?

Answer 42

ya

Answer 43

need almost 4 trapezium

Answer 44

where does the first trapezium end?

Answer 45

uhm..

Answer 46

end at 5

Answer 47

isit 5 ? o.o

Answer 48

yes 5, the second ends at 5+6 = 11

the third ends at 11+6 = 17

and so on...

Answer 49

since we need 3 whole trapezoids, we will start at x = 17 and add on fractional amounts of another whole trapezoid until we get to 31.5

Answer 50

31.5 - 24 = 7.5

we need just 7.5 square units of the 4th trapezium

Answer 51

We have this trapezoid

Answer 52

what is this area?