Question

Use the Intermediate Value Theorem to find intervals of length 1 which contain the real zeros of
f(x) = x3 − 9x + 5.

Answer 1

Answer on Question #55151 – Math – Algebra

Question

Use the Intermediate Value Theorem to find intervals of length 1 which contain the real zeros of $f(x) = x^3 - 9x + 5$ .

Solution

The first method is analytical with application of the Intermediate Value Theorem. First, just starting anywhere, $f(0) = 5 > 0$ . Next, $f(1) = -3 < 0$ . So, since $f(0) > 0$ and $f(1) < 0$ , there is at least one root in [0,1], by the Intermediate Value Theorem. Next, $f(2) = -5 < 0$ , $f(3) = 5 > 0$ . So, since $f(2) < 0$ and $f(3) > 0$ , by the Intermediate Value Theorem there is a root in [2,3]. Now if we somehow imagine that there is a negative root as well, then we try $-1$ : $f(-1) = 13 > 0$ . So we know nothing about roots in $[-1,0]$ . But continue: $f(-2) = 15 > 0$ , $f(-3) = 5 > 0$ and still no new conclusion. Continue: $f(-4) = -23 < 0$ . So, since $f(-3) > 0$ and $f(-4) < 0$ , by the Intermediate Value Theorem there is a third root in the interval $[-4, -3]$ .

Then, by the Intermediate Value Theorem, there are the zeros in the intervals [0,1], [2,3] and $[-4, -3]$ .

The second method is graphical. We search for points, where the graph crosses the x-axis.

Plot of the function $f(x) = x^3 - 9x + 5$ is given below.

There are the zeros in the intervals [0,1], [2,3] and $[-4, -3]$ .

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