PLEASE NOTE: Each * will indicate one full mark earned on solutions where full marks are not awarded.

Task 1: Knowledge/Understanding questions

7/11 marks

1. Evaluate the following limits algebraically

a. limx→3

4 x2+1

Answer

limx→3

4 x2+1

¿4 (3 )2+1¿4 (9 )+1¿36+1¿37 2*

b. limx→2

x2+2x−8x2−7 x+10

Answer

limx→2

x2+2x−8x2−7 x+10

need to factor then cancel then sub in f(2)

¿ x2+2 x−8x2−7x+10

¿22+2(2)−8

22−7(2)+10

¿ 4+4−84−14+10

¿ 8−8−10+10

¿0

c. limx→0

√ x+1−1x

Answer

¿ √x+1−1x

× √x+1+1√x+1+1

¿( x+1 )−1x(√x+1+1)

¿ xx(√x+1+1)

¿ 1√x+1+1

¿ 1√(0)+1+1

¿ 1√1+1

¿ 11+1

¿ 12 5*

3/3 marks2. Determine whether the limit exists as x approaches -2 for the following piecewise

function. Explain your answer.

f ( x )= {5x2−3 , x←2 } {4 x+25 , x>−2 }

Answer

x f ( x )=5 x2−3-3 42-2.5 28.25-2.01 17.20-2.001 17.02

x f ( x )=4 x+25-1 21-1.5 19

-1.9 17.4-1.99 17.04

As x approaches 2 from right-hand and left-hand side of the graph, the limit is 17. Since the limit is the same value when it is tested from both sides, the limit exists.Therefore limx→2

f ( x )=17

11/11 marks3. Using the rules of derivatives, find the derivative of each of the following

a. f ( x )=2 (x2−3 )(x3−5 x2)

Answer

f ( x )=2 (x2−3 ) (x3−5 x2 )f ' ( x )=(2 x2−6 ) (x3−5 x2 )f ' ( x )=( 4 x ) (x3−5 x2 )+(2 x2−6 ) (3 x2−10 x )f ' ( x )=4 x4−20 x3+6 x4−20 x3−18 x2+60xf ' ( x )=10 x4−40 x3−18 x2+60 x

b. g ( x )=3 (x3−5 x+1 )4

Answer

g ( x )=3 (x3−5 x+1 )4

g' (x )=3 [ 4 (x3−5x+1 )3 (3 x2−5 ) ]g' (x )=12 (3 x2−5 ) (x3−5 x+1 )3

c. y= 6 x2+15 x2−2

Answer

y= 6 x2+15 x2−2

dydx

=(12 x ) (5 x2−2 )−(6 x2+1 )(10 x)

(5 x2−2 )2

dydx

=60 x3−24 x−60 x3−10x(5 x2−2 )2

dydx

= −34 x( 5x2−2 )2

Task 2: Thinking questions

4/5 marks4. For the function f ( x )=2x3+3 x2, determine the points on the curve, where the slope of

the tangent is 36.

Answer

f ( x )=2x3+3 x2

f (a )=2a3+3a2

f (a+h )=2 (a+h )3+3 (a+h )2

f (a+h)=2 (a3+h3+3ah2+3a2h )+3 (a2+h2+2ah )f (a+h)=2a3+2h3+6ah2+6a2h+3a2+3h2+6 ah

Using the limit formula

limh→ 0

f (a+h )−f (a )h

limh→ 0

(2a3+2h3+6 ah2+6a2h+3a2+3h2+6ah )−(2a3+3a2)h

limh→0

(2a3+2h3+6 ah2+6a2h+3a2+3h2+6ah )−2a3−3 a2

h

limh→ 0

2h3+6ah2+6 a2h+3h2+6ahh

limh→ 0

h (2h2+6 ah+6a2+3h+6 a)h

limh→ 0

2h2+6 ah+6a2+3h+6 a

Substitute h=0¿2h2+6ah+6 a2+3h+6a¿2(0)2+6a (0)+6a2+3(0)+6 a¿0+0+6a2+0+6 a

¿6a2+6 a

Slope 6a2+6 a

We now need to find the tangent point for slope 36

36=6a2+6a6a2+6 a−36=0

x=−b±√b2−4ac2a

x=−6±√62−4 (6 )(−36)2(6)

x=−6±√36+86412

x=−6±√90012

x=−6±3012

a=2 and a=3

f (a )=2a3+3a2

f (2 )=2(2)3+3 (2 )2=28f (3 )=2(3)3+3(3)2=81 -27

The points on the curve where the slope of the tangent is 36 are (2,28 )∧(3,81) 4*

5/5 marks5. For the functions f ( x )=5 x3 and g ( x )=4 x2, prove that the derivative of the sum of the

function is equal to the sum of the derivative of each function.

[ f ( x )+g ( x ) ]'= f ' (x )+g' (x )

Answer

f ( x )=5 x3 and g ( x )=4 x2

f ( x )+g ( x )=5 x3+4 x2

[ f ( x )+g ( x ) ]'=15x2+8 x

f ' ( x )=15 x2

g' (x )=8 x

f ' ( x )+g' ( x )=15 x2+8 x15 x2+8x=15 x2+8 x

Hence [ f ( x )+g ( x ) ]'= f ' (x )+g' (x )

13/15 marks

6. Sketch the graph of y=−23x3+x2+24 x−5 by determining the location of any local

maximum or minimum points, and the intervals of increasing and decreasing.

Answer

As we know that the slope of a local maximum or minimum point is 0, we will first calculate the general formula of slopes and then find the value of the points by substitution.

y=−23x3+x2+24 x−5

f ( x )=−23x3+x2+24 x−5

f ( x+h )=−23

( x+h )3+( x+h )2+24 ( x+h )−5

f ( x+h )=−23

(x3+h3+3 x2h+3 xh2 )+ x2+h2+2 xh+24 x+24 h−5

f ( x+h )=−23x3−2

3h3−2 x2h−2 xh2+x2+h2+2xh+24 x+24h−5

f ' ( x ) limh→0

¿f ( x+h )−f (x )

h

¿

−23x3−2

3h3−2 x2h−2 x h2+x2+h2+2xh+24 x+24 h−5−(−2

3x3+x2+24 x−5)

h

¿

−23x3−2

3h3−2 x2h−2 x h2+x2+h2+2xh+24 x+24 h−5+ 2

3x3−x2−24 x+5

h

¿

−23h3−2 x2h−2x h2+h2+2 xh+24h

h

¿h(−23h2−2x2−2 xh1+h+2x+24)

h

¿−23h2−2 x2−2 xh1+h+2 x+24

h=0

¿−23

(0 )2−2 x2−2 x (0 )+ (0 )+2x+24

¿−2x2+2 x+24

x=−b±√b2−4ac2a

x=−2±√22−4 (−2 )(24)2(−2)

x=4 and x=−3

Hence the two local points are as follows

Maximum point

f ( 4 )=−23

(4 )3+ (4 )2+24 (4 )−5

¿−23

(64 )+16+96−5

¿64.3(4,64.3)

At the local maximum point, the direction of the curve changes from increasing to decreasing, from left to right.

Minimum point

f (−3 )=−23

(−3 )3+(−3 )2+24 (−3 )−5

¿−23

(−27 )+9−72−5

¿−50(−3 ,−50)

At the local minimum point, the direction of the curve changes from decreasing to increasing, from left to right.

We will first plot the two extrema and then draw the curves.

13* very good except you never stated the intervals of increase and decrease. Showing the graph is not enough.

Task 3: Communication questions

12/12 marks7. Define each of the following mathematical terms:

a. Extrema

Answer

The maximum and minimum values of a function on a graph are collectively called extrema. Global and local maximum and minimum points are together called extrema.

b. Derivative

Answer

The derivative of a function is the instantaneous rate of change or the slope of a tangent line at a specific point on the function. A derivative is denoted asf ' ( x ), pronounced as f prime of x.

c. Constant function

Answer

A constant function is f (a )=a where a, is a real number. The graph of such a function is a horizontal line with a slope 0. The output value of such a function is the same as the input value. There is no variable in the function.

d. First principle of derivatives

Answer

The first principle of derivatives is the formula used to calculate the derivative of a

function. f ' ( x )= limh→0

f ( x+h )−f (x)h It is a function itself.

e. Differentiation

Answer

The process of finding a derivative of a function is called differentiation.

f. Instantaneous rate of change

Answer

The instantaneous rate of change is the slope of a tangent to a curve. It is the change of one function in relation to another at a specific point on the graph.

5/5 marks8. What is the difference between a local minimum point and a global minimum point?

Sketch a function where a local minimum point is also the global minimum point.

Answer

A local minimum point is the point on a graph where the direction changes from decreasing to increasing. It is called a local point as it is not necessarily the lowest point on the graph.

A global minimum point is the lowest point of a function on the graph. It is also called the Absolute minimum point. It can be a local minimum point or can also be end point on the graph.

8/8 marks9. For the function f ( x )= (3x+2 )(2 x2−1):

a. Find the derivative using the product rule

Answer

f ( x )= (3x+2 ) (2x2−1 )Let g ( x )=(3 x+2 )Then g' (x )=3

Let h ( x )=(2 x2−1 )Then h' ( x )=4 x

f ' ( x )=3 (2 x2−1 )+4 x (3 x+2 )f ' ( x )=6 x2−3+12 x2+8xf ' ( x )=18 x2+8 x−3

b. Expand and simplify the function, then use the sum and difference rule to find the derivative.

Answer

f ( x )= (3x+2 ) (2x2−1 )f ( x )=6x3−3x+4 x2−2f ( x )=6x3+4 x2−3 x−2f ' ( x )=18 x2+8 x−3

c. What do you conclude from the results in parts (a) and (b)?

Answer

The result of parts (a) and (b) are the same. This concludes that the derivative of a function that is a product of two functions can be calculated using the product formula. In case there are more than two functions it needs to be simplified to two functions to apply the product rule.

It proves that if F ( x )=f ( x )g ( x ) then F ' ( x )=f ' ( x )g ( x )+f ( x )g ' ( x ) The result will be the same if the function is simplified into one function and then derivative is calculated.

This proves that the product rule and sum and difference rule all can be used to arrive at a derivative.

Task 4: Application questions

14/15 marks10. The equation M (t )=12−0.2 t2 models the mass of un-dissolved table salt in a glass of

water, where M (t) represents the mass of salt, in grams, and t represents time, in seconds.

a. When will all of the table salt be dissolved? Round your answer to two decimal points.

Answer

All the salt will be dissolved when the mass of the salt in water is 0

M (t )=12−0.2 t2

0=12−0.2 t 2

12=0.2t 2120.2

=t2

t 2=60t=7.745966692t=7.75secondsAll the salt will be dissolved at 7.75 seconds 2*

b. Use the secant method to find the average rate at which the salt is dissolving for the interval5.5<t<6.5.

Answer

First point x=5.5y=12−0.2(5.5)2

y=5.95( x , y )=(5.5,5 .95)

Second pointx=6.5y=12−0.2(6.5)2

y=3.55( x , y )=(5.5,3 .55)

Slope ¿ 3.55−5.956.5−5.51

¿−2.41

The average rate at which the salt is dissolving for the interval5.5<t<6.5 is 2.4g/secThe rate is decreasing, hence the negative sign. 3*

c. Use the method of first principles to find the rate at which the salt is dissolving at exactly 6 seconds. (Note : Using the correct method and showing steps in this solution will be worth marks)

Answer

M (t )=12−0.2 t2

OR

f (t)=12−0.2 t2

f (t+h )=12−0.2 ( t+h )2

f ( t+h )=12−0.2 (t 2+h2+2th )f ( t+h )=12−0.2 t2−0.2h2−0.4 thf ( t+h )=−0.2 t2−0.2h2−0.4 th+12

f ' (t )=limh→0

f ( t+h )−f (t)h

f ' (t )=limh→0

(−0.2 t2−0.2h2−0.4 th+12)−(12−0.2t 2)h

f ' (t )=limh→0

−0.2 t 2−0.2h2−0.4 th+12−12−0.2 t2

h

f ' (t )=limh→0

−0.2h2−0.4 thh

f ' (t )=limh→0

h (−0.2h−0.4 t)h

f ' (t )=limh→0

−0.2h−0.4 t

¿−0.2 (0 )−0.4 t¿−0.4 t

When t=6, Slope ¿−0.4 (6 )=−2.4

The salt is dissolving at exactly 6 seconds at 2.4g/sec 5*

d. When is the salt dissolving at a rate of 1.6g/sec?

Answer

Slope=−0.4 t-1.6=−0.4 tt=−4

As time cannot be negative (true but should tell you there is a calc error when this occurs) , the answer is 4 seconds.*

e. Use GeoGebra to sketch a graph of the function between the x- and y- intercepts. Refer to the video in lesson two on how to create graphs with restrictions on the domain and insert them into your assignment. Make sure to label your axes with title and give graph an overall title

Answer3*

7/ 10 marks11. Determine the equations of the tangent line to the graph of f ( x )=3 x (5 x2+1) that are

parallel to the line y=8x+9. Use GeoGebra to sketch a graph of the function, tangent lines, and the line y=8x+9

Answer

We will first find the slope of line y=8x+9y=mx+bwhere ,m=slope∧b= y interceptHence the slope is 8

We will now find the point of tangency with slope 8 to the graph of f ( x )=3 x (5 x2+1)

f ( x )=3 x (5 x2+1)f ' ( x )=3 (5 x2+1 )+10x (3 x )f ' ( x )=15 x2+3+30x2

f ' ( x )=45x2+3

We will now equate the slope of the line with the function of the derivative to find the point of tangency.

8=45 x2+3

5=45 x2

545

=x2

x=13

f ( 13 )=3 ( 1

3 )(5( 13 )

2

+1) and

f (−13 )

f ( 13 )=5

9+1=¿

y=149

The equation is y=m (x−x1 )+ y1

y=8(x−13 )+ 14

9

y=8x−83+ 14

9

y=8x−389 need to subtract in the numerator not add

5*

As the equation y=8x+9 intersects the graph the new equation parallel to it also intersects the graph. There is no tangent to the graph of f ( x )=3 x (5 x2+1) which is parallel to the line y=8x+9.

2* based on your equations