9.1 Locating roots by change of sign — Practise with help

1. $f(0) = -3$ and $f(2) = 8 - 4 - 3 = 1$.
2. Set $x^2 - 5 = 0$, so $x^2 = 5$ and $x = \pm\sqrt{5}$. The points are $(\sqrt{5}, 0)$ and $(-\sqrt{5}, 0)$.

If the first part was hard, revise substitution into a function in Section 2.9. If the second was hard, revise solving $x^2 = k$ in Section 2.4.

Write the left side as $f(x)$ and work out $f(1)$ and $f(2)$. One of them is negative and one is positive.

Compute $f(1) = 1 - 2 - 3 = -4$ and $f(2) = 8 - 4 - 3 = 1$.

Let $f(x) = x^3 - 2x - 3$. Then

The function is continuous and the signs differ, so the graph crosses the axis between $1$ and $2$. The root lies in $(1, 2)$.

The whole expression is already equal to zero, so use the left side as $f(x)$. Substitute $x = 1$ and $x = 2$.

Work out $f(1) = 1 + 2 - 5$ and $f(2) = 8 + 4 - 5$. One value is below zero and the other is above.

Let $f(x) = x^3 + 2x - 5$. Then

The signs are opposite, so a root lies in $(1, 2)$.

Bring every term to one side. The equation becomes $e^x - 4x = 0$, so the function is $f(x) = e^x - 4x$.

Work out $f(0) = 1 - 0 = 1$ and $f(1) = e - 4$. Use $e = 2.718\ldots$

Let $f(x) = e^x - 4x$. Then

The signs are opposite, so a root lies in $(0, 1)$.

Begin from the bracket $(1, 2)$ found in question 2. Test a value in the middle and keep the half that still contains the sign change.

Test $x = 1.5$: the value is $3.375 + 3 - 5 = 1.375$, which is positive. The root lies in $(1, 1.5)$. Then test $1.3$ and $1.4$.

The bracket is $(1, 2)$. Test $x = 1.5$:

so the root lies in $(1, 1.5)$. Test $x = 1.3$ and $x = 1.4$:

The root lies in $(1.3, 1.4)$. Test the midpoint, $x = 1.35$:

Now the sign changes between $1.3$ and $1.35$, so the root is below $1.35$ and rounds to $1.3$ to one decimal place.

Look at the value $x = 2$ and ask what happens to the function there.

The function is not defined at $x = 2$. Check whether $1/(x - 2)$ can ever equal zero.

The function has a break at $x = 2$, where it is undefined. Its graph jumps from large negative values to large positive values across the break, so the sign change comes from the break and not from a crossing. Also, a fraction $\dfrac{1}{x - 2}$ is never zero for any finite $x$, so the equation $\dfrac{1}{x - 2} = 0$ has no solution. The function is not continuous on $[1, 3]$, so the change-of-sign test does not apply.

A root needs a sign change between two consecutive readings in the table.

Compare the signs of $-1$, $-0.5$, $0.8$ and $4$ in order and find the first place where the sign changes.

The values at $x = 0$ and $x = 1$ are negative. The value at $x = 2$ is positive. The sign changes between $x = 1$ and $x = 2$, so a root lies in $(1, 2)$.

Work out the two values first. Both are positive.

A square is never negative, so the graph touches the axis at $x = 2$ and turns back. It does not cross.

Both values are positive, so there is no sign change on $(1, 3)$. The root at $x = 2$ is repeated: the graph meets the axis and turns back without crossing. A change-of-sign search misses a repeated root.