smc-737-60-Find f(x) given f'(x)

Consider the composite function `g(x)=f(\sin (2 x))`, where the function `f(x)` is an unknown but differentiable function for all values of `x`.

Use the following table of values for `f` and `f^{\prime}`.

`\quad x \quad`	`\quad\quad 1/2\quad\quad`	`\quad\quad(sqrt{2})/2\quad\quad`	`\quad\quad(sqrt{3})/2\quad\quad`
`f(x)`	`-2`	`5`	`3`
`\quad\quad f^{prime}(x)\quad\quad`	`7`	`0`	`1/9`

Find the value of `g\left(\frac{\pi}{6}\right)`. (1 mark)

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

The derivative of `g` with respect to `x` is given by `g^{\prime}(x)=2 \cdot \cos (2 x) \cdot f^{\prime}(\sin (2 x))`.

Show that `g^{\prime}\left(\frac{\pi}{6}\right)=\frac{1}{9}`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Find the equation of the tangent to `g` at `x=\frac{\pi}{6}`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Find the average value of the derivative function `g^{\prime}(x)` between `x=\frac{\pi}{8}` and `x=\frac{\pi}{6}`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Find four solutions to the equation `g^{\prime}(x)=0` for the interval `x \in[0, \pi]`. (3 marks)

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

Show Answers Only

a. `3`

b. `1/9`

c. `y=1/9x+3-pi/54`

d. `-48/pi`

e. ` x = pi/8 , pi/4 , (3pi)/8 ,(3pi)/4`

Show Worked Solution

a. `g(pi/6)`	`= f(sin(pi/3))`
	`= f(sqrt3/2)`
	`= 3`

b. `g\ ^{prime}(x)`	`= 2\cdot\ cos(pi/3)\cdot\ f\ ^{prime}(sin(pi/3))`
`g\ ^{prime}(pi/6)`	`= 2 xx 1/2 xx f\ ^{prime}(sqrt3/2)`
	`= 1/9`

c. `m = 1/9` and `g(pi/6) = 3`

`y – y_1`	`= m(x-x_1)`
`y – 3`	`= 1/9(x-pi/6)`
`y`	`= 1/9x + 3-pi/54`

♦♦ Mean mark (c) 45%.
MARKER’S COMMENT: Some students did not produce an equation as required.

d. The average value of `g^{\prime}(x)` between `x=\frac{\pi}{8}` and `x=\frac{\pi}{6}`

Average	`= \frac{1}{\frac{\pi}{6}-\frac{\pi}{8}}\cdot\int_{\frac{\pi}{8}}^{\frac{\pi}{6}} g^{\prime}(x) d x`
	`=24/pi \cdot[g(x)]_{\frac{\pi}{8}}^{\frac{\pi}{\6}}`
	`= 24/pi \cdot(f(sqrt3/2)-f(sqrt2/2))`
	`= 24/pi (3-5) = -48/pi`

♦♦ Mean mark (d) 30%.
MARKER’S COMMENT: Those who used the Average Value formula were generally successful.
Some students substituted `g^{\prime}(x)`, not `g(x)`.

e. `2 \cos (2 x) f^{\prime}(\sin (2 x)) = 0`

`:.\ 2 \cos (2 x) = 0\ ….(1)` or ` f^{\prime}(\sin (2 x)) = 0\ ….(2)`

(1): ` 2 \cos (2 x)`	`= 0`	`x \in[0, \pi]`
`\cos (2 x)`	`= 0`	`2 x \in[0,2 \pi]`
`2x`	`= pi/2 , (3pi)/2`
`x`	`= pi/4 , (3pi)/4`

(2): ` f^{\prime}(\sin (2 x)) `	`= sqrt2/2`
`2x`	`= pi/4 , (3pi)/4`
`x`	`= pi/8 , (3pi)/8`

`:. \ x = pi/8 , pi/4 , (3pi)/8 ,(3pi)/4`

♦♦ Mean mark (e) 30%.
MARKER’S COMMENT: Some students were able to find `pi/4, (3pi)/4`. Some solved `2 cos(2x)=0` or `f^{\prime}(sin(2x))=0` but not both.

`int f(x)\ dx`	`=int cos\ 3x\ dx`
	`= 1/3 sin\ 3x + c`
`f(pi/6)`	`= 1/3\ [sin\ (3 xx pi/6) ]+ c`
`1`	`= 1/3\ sin\ pi/2+c`
`c`	`= 2/3`

`1/pi`	`= (-cos (2 pi))/(2 pi) + c`
`1/pi`	`= -1/(2 pi) + c`
`:. c`	`= 3/(2 pi)`

`1/2`	`= 2sin(pi/2) + 1/2cos(pi) + c`
`1/2`	`= 2-1/2 + c`
`c`	`=-1`
`:. f(x)`	`= 2sinx + 1/2cos(2x)-1`

Calculus, MET2 2022 VCAA 5

Calculus, MET1 SM-Bank 5

Calculus, MET1 2013 VCAA 3

Calculus, MET1 2014 VCAA 7

`g^{prime} (x)`	`= int sin (2 pi x) dx`
	`= (-cos (2 pi x))/(2 pi) + c`

`f(x)`	`= int(2cosx-sin2x)dx`
	`= 2sinx + 1/2cos(2x) + c`