Locating roots by change of sign

Use the substitution u = cos x to show that

Fig. 1 shows part of the curve y = The shaded region is bounded by the curve, the x-axis, and the line x = a.

You are given that the area of the shaded region is a square units.

Show that the value of a satisfies the equation a - cos^-1(1 - 2e^-a) = 0.

Show by calculation that the value of a lies between 1.1 and 1.2.

Use the iterative formula

a_n+1 = cos^-1(1 - 2e^-a_n),

with starting value a₁ = 1.2, to find the value of a correct to 2 decimal places. Show the

result of each step of the iterative process.

Fig. 2 shows the curve y = cos^-1(1 - 2e^-x).

Explain why the gradient of the curve y = cos^-1(1 - 2e^-x) at the point where x = a, where a

is the value found in part (d), lies in the interval (- 1, 0).

The diagram shows part of the curve f(x) = + 2. The equation f(x) = 0 has a positive root α close to x = 0.3.

Explain why using the sign change method with x = 0 and x = 1 will fail to locate α.

Show that the equation f(x) = 0 can be written as x =

An alternative iterative formula is x_n+1 = F(x_n), where F(x_n) = ln(2 - 8x_n²).

By considering F ′ (0.3) explain why this iterative formula will not find α.

In this question you must show detailed reasoning.

The diagram shows the curve with equation y = . The tangent to the curve at the point P has gradient 2.

Show that the x-coordinate of P satisfies the equation

4x³ + 3x - 3 = 0.

Show by calculation that the x-coordinate of P lies between 0.5 and 1.

Show that the iteration

cannot converge to the x-coordinate of P whatever starting value is used.

Use the Newton-Raphson method, with initial value 0.5, to determine the coordinates of P

correct to 5 decimal places.

The diagram shows part of the graph of y = f(x), where f(x) is a cubic polynomial in x.

Explain why one of the roots of the equation f(x) = 0 cannot be found by the sign change method.

The equation 6 arcsin(2x - 1) - x² = 0 has exactly one real root.

Show by calculation that the root lies between 0.5 and 0.6.

In order to find the root, the iterative formula

x_n+1 = p + q sin(rx_n²),

with initial value x₀ = 0.5, is to be used.

Determine the values of the constants p, q and r.

Hence find the root correct to 4 significant figures. Show the result of each step of the iteration

process.

A curve has equation y = 2 ln(k - 3x) + x² - 3x, where k is a positive constant.

Given that the curve has a point of inflection where x = 1, show that k = 6.

It is also given that the curve intersects the x-axis at exactly one point.

Show by calculation that the x-coordinate of this point lies between 0.5 and 1.5.

Use the Newton-Raphson method, with initial value x₀ = 1, to find the x-coordinate of the point where the curve intersects the x-axis, giving your answer correct to 5 decimal places.

Show the result of each iteration to 6 decimal places.

By choosing suitable bounds, verify that your answer to part (c) is correct to 5 decimal places.

Show that θ = 0.4 + sin θ.

Let F (θ) = 0.4 + sin θ.

By considering the value of F ′ (θ) where θ = 1.2, explain why using an iterative method based on the equation in part (a) will converge to the root, assuming that 1.2 is sufficiently close to

the root.

Use the iterative formula θ_n+1 = 0.4 + sin θ_n with a starting value of 1.2 to find the value of θ correct to 4 significant figures.

You should show the result of each iteration.

Use a change of sign method to show that the value of θ found in part (c) is correct to 4

significant figures.