Question

Math question, please help.

Answer 1

First, let's use the Intermediate Value Theorem (IVT) to show that a zero exists within the region of [0, 1]. To do this, we are going to evaluate the function at x = 0 and x = 1. If one of the evaluations is negative and the other is positive, we know that some $c$ exists where $f(c) = 0$.

$f(0) = 7\cos(0) - 12(0) = 7$

$f(1) = 7\cos(1) - 12(1) = 7\cos(1) - 12 \approx -8.217$

We have successfully proved that there is a 0 within the interval of [0, 1]. Now, let's use the bisection method to find it. The problem is asking us for an interval with length 1/32, which means we are going to have to do the bisection method five times.

The reasoning for doing it five times is as such:

We are going to do the bisection method $n$ times, which gives us an interval of length $\frac{b - a}{2^n}$ we are splitting the interval [a, b] in half with each iteration). In the case of this problem, where $\frac{1}{32} = \frac{1}{2^5}$, which means we are going to do the bisection method five times.)

Now, let's start doing the bisection method. To do the bisection method, here are the steps:

1. Evaluate the mid-point of your interval ($\frac{a + b}{2}$)
2. If the evaluation is positive, create the interval of the mid-point and the endpoint of the previous interval which was negative. If the evaluation is zero, you are done. If the evaluation is negative, then create an interval of the mid-point and the endpoint of the previous interval which was positive.
3. Repeat until satisfied.

The first interval is [0, 1], meaning our mid-point is $\frac{1}{2}$. Testing our mid-point, we get:

$f(\frac{1}{2}) = 7\cos(\frac{1}{2}) - 12(\frac{1}{2}) = 7\cos(\frac{1}{2}) - 6 \approx 0.143$

Since our evaluation was positive, our new interval will be [$\frac{1}{2}$, 1].

Our mid-point for this interval is $\frac{3}{4}$. Testing, we get:

$f(\frac{3}{4}) = 7\cos(\frac{3}{4}) - 12(\frac{3}{4}) \approx -3.878$

Our next interval will be [$\frac{1}{2}$, $\frac{3}{4}$]

Mid-point of [$\frac{1}{2}$, $\frac{3}{4}$]: $\frac{5}{8}$

$f(\frac{5}{8}) \approx -1.823$

Next interval: [$\frac{1}{2}$, $\frac{5}{8}$]

Mid-point of [$\frac{1}{2}$, $\frac{5}{8}$]: $\frac{9}{16}$

$f(\frac{9}{16}) \approx -0.823$

Next interval: [$\frac{1}{2}$, $\frac{9}{16}$]

The interval of [$\frac{1}{2}$, $\frac{9}{16}$] is the smallest interval we can find that is an answer choice, which is the first answer choice. To prove that this is the answer, solving by graphing shows that the zero is within this interval, $x \approx 0.51$.