The value of the gradient of f(x) =-2sinx. At every /4 unit is shown in the table below:
The scatter plots of values in the table is sketched below: (joined by a smooth curve)
gradient of f(x) =-2sinx at every /4 unit
③ when a=2,f(x) =2sinx.
The value of the gradient of f(x) =2sinx. At every /4 unit is shown in the table below:
The scatter plots of values in the table is sketched below: (joined by a smooth curve)
gradient of f(x) =2sinx at every /4 unit
b) Make a conjecture of g′(x)
g′(x)=a cosx
c) Test the conjecture with further examples.
1. Let a=3 , then g′(x)=3 cosx.
The value of the gradient of f(x) =3sinx. at every /4 unit is shown in the table below:
The scatter plot for the values in the table above with the function g′(x)=3 cosx. is shown in figure 4 The derivative of f(x) =3sinx found by the calculator with the curve of g′(x)=3 cosx is shown in figure 5
Figure 4 g′(x)=3 cosx. Figure 5 The derivative of f(x) =3sinx
Figure 4 reveals the curve of g′(x)=3 cosx fits the scatter plot very well. Figure 5 indicates that the derivative function found by calculator overlaps g′(x)=3 cosx. Perfect, they should be the same function. Therefore g′(x)=a cosx is the correct conjecture for function g(x)=asinx.
2. Method and findings of the derivative of f(x) =asinx:
The curves of conjecture, when a=-1, a=-2and a=2, in the graphs indicate the following characteristics:
·symmetrical along x-axis.
·repeat its values every 2,(so the function is periodic with a period of 2)
·the maximum value is a, the minimum value is -a
·the mean value of the function is zero
·the amplitude of the function is a
According to the characteristics the form of gradient function is similar to a cosine function with parameter of a. Therefore the conjecture of g’(x) is: g′(x)=a cosx.
d) State for what values of a the conjecture holds/
For all values of a the conjecture holds.
The letter a determines the minimum, maximum values and amplitude of sine,cosine function.
When a>0 the curve will be vertical stretch by a factor of a.
Say a=2 g(x)=2sinx , g′(x)=2cosx (graph is shown below)
When a<0 the curve will be reflection along x-axis and then vertical stretch by a factor of a
Say a=-3 g(x)= -3sinx ,g′(x)= -3cosx (graph is shown below)
When a=0 g(x)= 0 , g′(x)= 0 the curve is still exist as y=0.(graph is shown below)
g(x)=2sinx ,g′(x)=2cosx g(x)= -3sinx, g′(x)= -3cosx g(x)= 0, g′(x)= 0
As a result a can be positive, negative or zero, so for all values of a the conjecture holds.
Question 3
Investigate the derivatives of functions of the form f(x) =sin(x+c).
a) Consider several different values of c.
1.Consider c=1, c =2 and c =-1
① when c =1: f(x) =sin(x+1) ②when c =-1 f(x) =sin(x-1) ③when c =2 f(x) =sin(x+2)
the graph is shown below:
graph of f(x)=sin(x+1),f(x)=sin(x-1),
f(x) =sin(x+2)
2. Description the gradient of the function:
① when c=1: f(x) =sin(x+1)
·The line of the tangent becomes flatter and flatter as the points move from left to right within -2to--1,--1to--1,-1to-1,-1to-1and 2-1to 2. The line of the tangent become steeper and steeper as the points move from left to right within--1 to --1,--1 to -1,-1 to -1,-1to 2-1 .
·The gradient is positive in the domain of:[-2,--1[,]--1, -1[,
]-1, 2]. The gradient is negative in the domain of::]--1,--1[,
]-1, -1[.
② when c =-1 f(x) =sin(x-1)
·The line of the tangent becomes flatter and flatter as the points move from left to right within
-2+1 to -+1,-+1 to -+1,1to +1,+1 to +1.
The line of the tangent becomes steeper and steeper as the points move from left to right within-2to-2+1,-+1to-+1,-+1 to 1,+1 to +1,+1to 2.
·The gradient is positive in the domain of :[-2,-+1[, ]-+1, +1[,
]+1,2]
The gradient is negative in the domain of::]-+1,-+1[, ]+1, +1[.
③when c =2 f(x) =sin(x+2)
·The line of the tangent becomes flatter and flatter as the points move from left to right within
--2to --2,-2to-2,-2 to -2,2-2 to -2.
The line of the tangent becomes steeper and steeper as the points move from left to right within-2to --2,--2to-2,-2to-2, -2to 2-2, -2 to 2.
·The gradient is positive in the domain of :]--2,-2[,]-2, -2]
The gradient is negative in the domain of:]-2,--2[, ]-2,-2[
]-2, 2].
3. Values of gradient of the function at every /4 unit, sketch findings on a graph.
①The value of the gradient of f(x) =sin(x+1) at every /4 unit is shown in the table below:
The scatter plots of values in the table is sketched below: (joined by a smooth curve)
gradient of f(x) =sin(x+1) at every /4 unit
②The value the gradient of f(x) =sin(x-1) at every /4 unit is shown in the table below:
The scatter plots of values in the table is sketched below: (joined by a smooth curve)
gradient of f(x) =sin(x-1) at every /4
③The value of the gradient of f(x) =sin(x+2) at every /4 unit is shown in the table below:
The scatter plots of values in the table is sketched below: (joined by a smooth curve)
gradient of f(x) =sin(x+2) at every /4
b) Make a conjecture of g′(x)
g′(x)=cos(x+ c)
c) Test the conjecture with further examples.
consider c =-2 then f(x) =sin (x-2), g′(x)=cos(x-2)
The value of the gradient of f(x) =sin(x-2) at every /4 unit is shown in the table below:
The scatter plot for the values in the table above with the function g′(x)=cos(x-2) is shown in figure 6. The derivative of f(x) =sin(x-2) found by the calculator with the curve of g′(x)=cos(x-2) is shown in figure 7:
Figure 6 g′(x)=cos(x-2) figure 7 derivative of f(x) =sin(x-2)
Figure 6 reveals the curve of g′(x)=cos(x-2) fits the scatter plot well. Figure 7 reveals that the derivative function found by calculator overlaps g′(x)=cos(x-2). Perfect.
Therefore g′(x)=cos(x+ c) is the suitable conjecture for function f (x)=sin(x+ c).
2. Method and findings of the derivative of f (x)=sin(x+ c)
The curves of conjecture, when c =-1, c=1and c=2, in the graphs indicate the following characteristics:
·symmetrical along the axis of x-c.
·repeat its values every 2,(so the function is periodic with a period of 2)
·the maximum is 1 the minimum is -1
·the mean value of the function is zero
·the amplitude of the function is 1
According to the characteristics the form of function for the gradient is similar to a cosine function with parameter of 1 and horizontal transformed by c units to the left.
Therefore correct onjecture of g’(x) is: g′(x)=cos(x+c)
d) State for what values of c the conjecture holds.
For all values of c the conjecture holds.
Because c won’t affect any characteristics of the function.
When c>0 it will cause horizontal translation to the right by c units.
When c<0 it will cause horizontal translation to the left by c units.
When c =0 the curve will be the same as f (x)=sinx, g′(x)=cosx.
So c can be positive, negative or zero it means for any value of c the conjecture holds.
Question 4
Investigate the derivatives of functions of the form f(x) =sinbx
a) Consider several different values of b.
1.Consider b=0.5, b =-0.5 and b =-1 When b =-1 f(x) =sin-x , when b =0.5 f(x) =sin0.5x when b =-0.5 f(x) =sin-0.5x the graph is shown below:
graph of f(x) =sin-x f(x) =sin0.5x .f(x) =sin-0.5x
①When b =-1 f(x) =sin-x
The behaviour of the gradient of the function on the given domain is opposite as question 1 (b).
② when b =0.5 the f(x) =sin0.5x
·The line of the tangent becomes flatter and flatter as the points move from left to right within
-2 to -, 0to .
The line of the tangent becomes more and more precipitous as the points move from left to right within- to 0,to 2.
·The gradient is positive in the domain of : ]-,[
The gradient is negative in the domain of::[-2,-[, ],2]
③when b =-0.5 the f(x) =sin-0.5x
·The line of the tangent becomes flatter and flatter as the points move from left to right within
-2 to -, 0to .
The line of the tangent becomes more and more precipitous as the points move from left to right within - to 0,to 2.
·The gradient is positive in the domain of : [-2,-[, ],2]
The gradient is negative in the domain of:: ]-,[
3. Values of gradient of the function at every /4 unit, sketch findings on a graph.
①The value of the function of f(x) =sin-x. at every /4 unit is shown in the table below:
The scatter plots of values in the table is sketched below: (joined by a smooth curve)
gradient of f(x) =sin-x. at every /4 unit
②The value the gradient of f(x) =sin0.5x at every /4 unit is shown in the table below:
The scatter plots of values in the table is sketched below: (joined by a smooth curve)
gradient of f(x) =sin0.5x at every /4 unit
③The value the gradient of f(x) =sin0.5x at every /4 unit is shown in the table below:
The scatter plots of values in the table are sketched below: (joined by a smooth curve):
gradient of f(x) =sin-0.5x at every /4 unit
b) Make a conjecture of g′(x)
g′(x)=bcosbx
c) Test the conjecture with further examples.
1.consider b=2 then f(x) =sin2x, g′(x)=2cos2x
The value of the gradient of f(x) =sin2x at every /4 unit is shown in the table below:
The scatter plot for the values in the table above with the function g′(x)= 2cos2x is shown in figure 8. The derivative of f(x) = sin2x found by the calculator with the curve of g′(x)= 2cos2x is shown in figure 9:
Figure 8 g′(x)= 2cos2x figure 9 derivative of f(x) = sin2x
Figure 8 reveals the curve of g′(x)=2cos2x fits the scatter plot well. Figure 9 reveals that the derivative function found by calculator overlaps g′(x)=cos(x-2). Perfect.
Therefore g′(x)= bcosbx is the suitable conjecture for function f (x)=sinbx.
2. Method and findings of the derivative of f (x)=sinbx
The curves of conjecture, when b=-1, b=0.5and b=-0.5, in the graphs reveal the following characteristics:
·symmetrical along the x-axis.
·repeat its values every ,(so the function is periodic with a period of )
·the maximum is b the minimum is -b
·the mean value of the function is zero
·the amplitude of the function is b
According to the characteristics the form of function for the gradient is similar to a cosine function with parameter of b and horizontal stretched by a factor of .
Therefore the correct onjecture of g’(x) is: g′(x)= bcosbx.
d) State for what values of b the conjecture holds.
For all values of b the conjecture holds.
Because b determines the period of sine,cosine function.
No matter b is positive or negative; the curve will be horizontal stretched by a factor of
When b=0, f(x) =0 the conjecture still exists as: g′(x)=0.
Therefore for all values of b the conjecture holds.
Question 5
Use your results in Question 1 to 4 to make a conjecture for the derivative of k(x)=asinb(x+c).
Choose a value for each of a,b,c. Verify your conjecture using the values you have chosen for a, b and c.
-
k(x)=asinb(x+c)
conjecture: k′(x)= abcosb(x+ c)
-
1. let a=-1, b=0.5, c=1,Then k(x)= -sin0.5(x+1), k′(x)= -0.5cos0.5(x+1)
2. The value of the gradient of f(x) =sin2x at every /4 unit is shown in the table below:
c) The scatter plot for the values in the table above with the function k′(x)= -0.5cos0.5(x+1) is shown in figure 10. The derivative of k(x)= asinb (x+ c) found by the calculator with the curve of k′(x)= -0.5cos0.5(x+1) is shown in figure 11:
Figure 10 k′(x)= -0.5cos0.5(x+1) figure 11 derivative of k(x)= asinb (x+ c)
Figure 10 reveals the curve of k′(x)= -0.5cos0.5(x+1) fits the scatter plot well. Figure 11 reveals the derivative function found by calculator overlaps k′(x)= -0.5cos0.5(x+1) Perfect.
Therefore k′(x)= -0.5cos0.5(x+1) is the correct onjecture for function k(x)=asinb (x+ c)
Question 6
Consider m(x)=sinx. Investigate the derivative of this function and show that it can be written as m′(x)=2sinxcosx.
-
Graph of the function m(x)=sinx ,For-2x2
y=m(x),let m(x)=sinx ,-2x2
By sketching the graph we get:
From the graph we can get the range of the function m(x)=sinx. is [0,1].
b) Describe the behaviour of the gradient of the function on the given domain.
According to the graph the gradient of the function reveals these characteristics:
·The line of the tangent becomes flatter and flatter as moving from left to right within
-2 to -,-to -,- to-, to -,0to, to , to , to .
The line of the tangent becomes more and more precipitous as moving from left to right within - to -, - to -,- to -, - to 0, to , to , to , to 2.
·In the domain of : ]-2, -],]-, -[, ]0,[ , ],[ the gradient is positive. In the domain of::]-,-[, ]-,0[, ],[, ], 2[ the gradient is negative
·At the point when x equals to -2,-,-,-,0,, , and 2, the gradient is obviously 0.
c) Find values of gradient of the function at /4 every unit and sketch findings on a graph
The numerical values of the gradient of the function at every /4 unit is showed in the table:
The scatter plot for the gradient of f(x) =sinx. at every /4 unit is shown below.
gradient of f(x) =sin(x+1) at every /4
d) Make a conjecture for the derive function m′(x)
the conjecture for the derived function is m′(x)=sin2x.
e) The scatter plot for the values in the table above with the function m′(x)=sin2x is shown in figure 12. The derivative of m(x)=sinx found by the calculator with the curve of
m′(x)=sin2x is shown in figure 13:
Figure 12 m′(x)=sin2x figure 13 derivative of m(x)=sinx
Figure 12 reveals the curve of m′(x)=sin2x fits the scatter plot well. Figure 13 reveals that the derivative function found by calculator overlaps m′(x)=sin2x). Perfect.
Therefore m′(x)=sin2x is the correct conjecture for function m(x)=sinx.
2. Method and findings:
The curve in figure 16 reveals the following characteristics:
·symmetrical along the axis of .
·repeat its values every ,(so the function is periodic with a period of )
·the maximum is 1 the minimum is -1
·the mean value of the function is zero
·the amplitude of the function is 1
According to the characteristics the form of function for the gradient is similar to a cosine function with parameter of 1 and horizontal stretched by units.
Therefore conjecture of m′(x) is: m′(x)=sin2x
Based on the double angle formula: sin2x=2sinxcosx.
the conjecture m′(x)=sin2x,
Combine the two equation together we get: m′(x)=sin2x=2sinxcosx.
Therefore m′(x)=sin2x is the same as m′(x)= 2sinxcosx
So the derivative of functio m(x)=sinx is m′(x)=sin2x and it can be written as
m′(x)=2sinxcosx.