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Iteration
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35 questions at your level
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Level 3
Level 4
Level 5
Showing a root exists — change of sign
13 questions
Lesson
Not started
Mark as done
f
(
x
)
=
x
3
+
x
−
20
f(x) = x^3 + x - 20
f
(
x
)
=
x
3
+
x
−
20
The equation
f
(
x
)
=
0
f(x) = 0
f
(
x
)
=
0
has a solution close to
x
=
2.6
x = 2.6
x
=
2.6
Show that this solution is
x
=
2.59
x = 2.59
x
=
2.59
, correct to 2 decimal places.
●●●●●
Level 5
3 marks
Start
→
Mark as done
f
(
x
)
=
x
3
−
4
x
−
9
f(x) = x^3 - 4x - 9
f
(
x
)
=
x
3
−
4
x
−
9
The equation
f
(
x
)
=
0
f(x) = 0
f
(
x
)
=
0
has a solution close to
x
=
2.7
x = 2.7
x
=
2.7
Show that this solution is
2.7
2.7
2.7
correct to 1 decimal place
.
●●●●●
Level 5
3 marks
Start
→
Using an iterative formula
15 questions
Lesson
Not started
Mark as done
x
n
+
1
=
7
x
n
+
8
3
x_{n+1} = \sqrt[3]{7x_n + 8}
x
n
+
1
=
3
7
x
n
+
8
Starting with
x
1
=
3
x_1 = 3
x
1
=
3
, the values of
x
n
x_n
x
n
get closer and closer to
3.0958
…
3.0958\ldots
3.0958
…
(a)
Write down a solution, correct to 2 decimal places, of the equation
x
3
−
7
x
−
8
=
0
x^3 - 7x - 8 = 0
x
3
−
7
x
−
8
=
0
(b)
Explain why this value is a solution of the equation
x
3
−
7
x
−
8
=
0
x^3 - 7x - 8 = 0
x
3
−
7
x
−
8
=
0
●●●●●
Level 5
2 marks
Start
→
Mark as done
x
n
+
1
=
4
x
n
+
9
3
x_{n+1} = \sqrt[3]{4x_n + 9}
x
n
+
1
=
3
4
x
n
+
9
(a)
Given that
x
3
=
2.68168
x_3 = 2.68168
x
3
=
2.68168
, work out the value of
x
4
x_4
x
4
. Give your answer correct to 3 decimal places.
(b)
The values of
x
n
x_n
x
n
approach a limit. Which equation does that limit satisfy?
●●●●●
Level 5
4 marks
Start
→
Mark as done
x
n
+
1
=
30
−
2
x
n
3
,
x
1
=
3
x_{n+1} = \sqrt[3]{30 - 2x_n}, \qquad x_1 = 3
x
n
+
1
=
3
30
−
2
x
n
,
x
1
=
3
(a)
Work out the values of
x
2
x_2
x
2
and
x
3
x_3
x
3
. Give your answers correct to 4 decimal places.
(b)
Explain what the limit of this iteration represents.
●●●●●
Level 5
4 marks
Start
→
Rearranging an equation into iterative form
13 questions
Lesson
Not started
Mark as done
(a)
Show that the equation
x
3
−
5
x
+
1
=
0
x^3 - 5x + 1 = 0
x
3
−
5
x
+
1
=
0
can be rearranged to give
x
=
x
3
+
1
5
x = \dfrac{x^3 + 1}{5}
x
=
5
x
3
+
1
(b)
Using
x
n
+
1
=
x
n
3
+
1
5
x_{n+1} = \dfrac{x_n^3 + 1}{5}
x
n
+
1
=
5
x
n
3
+
1
with
x
1
=
0.2
x_1 = 0.2
x
1
=
0.2
, work out the value of
x
2
x_2
x
2
●●●●●
Level 5
3 marks
Start
→
Mark as done
(a)
Show that the equation
x
3
−
4
x
−
9
=
0
x^3 - 4x - 9 = 0
x
3
−
4
x
−
9
=
0
can be rearranged to give
x
=
x
3
−
9
4
x = \dfrac{x^3 - 9}{4}
x
=
4
x
3
−
9
(b)
Using the iteration
x
n
+
1
=
4
x
n
+
9
3
x_{n+1} = \sqrt[3]{4x_n + 9}
x
n
+
1
=
3
4
x
n
+
9
with
x
1
=
2
x_1 = 2
x
1
=
2
, work out
x
2
x_2
x
2
. Give your answer correct to 4 decimal places.
●●●●●
Level 5
4 marks
Start
→