smc-1089-10-Graph f(x) given f'(x)

Let `y=f(x)` be a function defined for `0 <= x <= 6`, with `f(0)=0`.

The diagram shows the graph of the derivative of `f`, `y = f prime (x)`.

The shaded region `A_1` has area `4` square units. The shaded region `A_2` has area `4` square units.

For which values of `x` is `f(x)` increasing? (1 mark)

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What is the maximum value of `f(x)`? (1 mark)

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Find the value of `f(6)`. (1 mark)

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Draw a graph of `y =f(x)` for `0 <= x <= 6`. (2 marks)

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Show Answers Only

`f(x)\ text(is increasing when)\ 0 <= x < 2`
`text(MAX value of)\ f(x) = 4`
`-6`

Show Worked Solution

i.	`f(x)\ text(is increasing when)\ \ f prime (x) > 0`
	`text(From the graph)`
	`f(x)\ text(is increasing when)\ 0 <= x < 2`

ii.	`f prime (x) = 0\ \ text(when)\ \ x=2`
	`:.\ text(MAX at)\ \ x = 2`
	`int_0^2 f prime (x)\ dx = 4\ \ \ (text(given since)\ A_1 = 4 text{)}`
	`text(We also know)`

`int_0^2\ f prime (x)\ dx`	`= [f(x)]_0^2`
	`= f(2)\ – f(0)`
	`= f(2)\ \ \ \ text{(since}\ f(0) = 0 text{)}`

`=> f(2) = 4`

♦♦♦ Parts (ii) and (iii) proved particularly difficult for students with mean marks of 12% and 11% respectively.

`:.\ text(MAX value of)\ \ f(x) = 4`

iii.	`int_0^4 f prime (x)`	`= A_1\ – A_2`
		`=0`

`text(We also know)`

`int_0^4 f prime (x)\ dx`	`= int_2^4 f prime (x)\ dx + int_0^2 f prime (x)\ dx`
	`=[f(x)]_2^4 + 4`
	`= f(4)\ – f(2) + 4\ \ \ (text(note)\ f(2)=4)`
	`=f(4)`

`=> f(4) = 0`

`text(Gradient = – 3 from)\ \ x = 4\ \ text(to)\ \ x = 6`

`:.\ f(6)`	`= -3 (6\ – 4)`
	`= -6`

♦♦ Mean mark 28%
EXAM TIP: Clearly identify THE EXTREMES when given a defined domain. In this case, the origin is obvious graphically, and the other extreme at `x=6`, is CLEARLY LABELLED!

iv.