smc-7133-30-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)

--- 2 WORK AREA LINES (style=lined) ---
What is the maximum value of `f(x)`? (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Find the value of `f(6)`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Draw a graph of `y =f(x)` for `0 <= x <= 6`. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `f(x)\ text(is increasing when)\ 0 <= x < 2`

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

c. `-6`

Show Worked Solution

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

b.	`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 (b) and (c) proved particularly difficult for students with mean marks of 12% and 11% respectively.

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

c.	`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\ text(from)\ \ x = 4\ \ text(to)\ \ x = 6`

`:.\ f(6)`	`=-3 (6- 4)`
	`=-6`

♦♦ Mean mark (d) 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!