The equation $x^3 - 3x + 1 = 0$ has three real roots - AQA - A-Level Maths Mechanics - Question 4 - 2017 - Paper 2
Question 4
The equation $x^3 - 3x + 1 = 0$ has three real roots.
(a) Show that one of the roots lies between -2 and -1.
(b) Taking $x_1 = -2$ as the first approximation to on... show full transcript
Worked Solution & Example Answer:The equation $x^3 - 3x + 1 = 0$ has three real roots - AQA - A-Level Maths Mechanics - Question 4 - 2017 - Paper 2
Step 1
Show that one of the roots lies between -2 and -1.
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Answer
To determine if there is a root between -2 and -1, we will evaluate the function at these points:
Calculate f(−2):
f(−2)=(−2)3−3(−2)+1=−8+6+1=−1
Calculate f(−1):
f(−1)=(−1)3−3(−1)+1=−1+3+1=3
Since f(−2)=−1<0 and f(−1)=3>0, we observe that the function changes sign between -2 and -1.
By the Intermediate Value Theorem, since f(x) is continuous, there must be at least one root in the interval (−2,−1).
Step 2
Taking $x_1 = -2$ as the first approximation to one of the roots, use the Newton-Raphson method to find $x_2$, the second approximation.
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Answer
The Newton-Raphson method formula is given by:
xn+1=xn−f′(xn)f(xn)
First, compute the derivative of the function:
f′(x)=3x2−3
Substitute x1=−2 into both f(x) and f′(x):
f(−2)=−1 (as computed earlier)
f′(−2)=3(−2)2−3=12−3=9
Now apply the Newton-Raphson formula:
x2=−2−9−1=−2+91=−2+0.1111=−1.8889
Thus, the second approximation x2 is approximately −1.8889.
