Question

for the given function, determine consecutive values of x between which each real zero is located.

1. $f(x)=x^{3}-3x + 1$
2. $f(x)=2x^{3}+9x^{2}+12x + 2$

Explanation:

Step1: Evaluate function at integer values

We will evaluate \(f(x)=x^{3}-3x + 1\) at consecutive integer - values of \(x\). Let's start with \(x=-2\):
\[f(-2)=(-2)^{3}-3\times(-2)+1=-8 + 6+1=-1\]

Step2: Evaluate at \(x=-1\)

\[f(-1)=(-1)^{3}-3\times(-1)+1=-1 + 3+1=3\]
Since \(f(-2)=-1\lt0\) and \(f(-1)=3\gt0\), by the Intermediate - Value Theorem, there is a zero between \(x=-2\) and \(x=-1\).
Next, let \(x = 0\):
\[f(0)=0^{3}-3\times0 + 1=1\]
Let \(x = 1\):
\[f(1)=1^{3}-3\times1 + 1=1-3 + 1=-1\]
Since \(f(0)=1\gt0\) and \(f(1)=-1\lt0\), there is a zero between \(x = 0\) and \(x = 1\).
Now, for \(f(x)=2x^{3}+9x^{2}+12x + 2\). Let \(x=-3\):
\[f(-3)=2\times(-3)^{3}+9\times(-3)^{2}+12\times(-3)+2=2\times(-27)+9\times9-36 + 2=-54 + 81-36 + 2=-7\]
Let \(x=-2\):
\[f(-2)=2\times(-2)^{3}+9\times(-2)^{2}+12\times(-2)+2=2\times(-8)+9\times4-24 + 2=-16 + 36-24 + 2=-2\]
Let \(x=-1\):
\[f(-1)=2\times(-1)^{3}+9\times(-1)^{2}+12\times(-1)+2=-2 + 9-12 + 2=-3\]
Let \(x = 0\):
\[f(0)=2\times0^{3}+9\times0^{2}+12\times0+2=2\]
Since \(f(-1)=-3\lt0\) and \(f(0)=2\gt0\), there is a zero between \(x=-1\) and \(x = 0\).

Answer:

For \(f(x)=x^{3}-3x + 1\), zeros are between \(x=-2\) and \(x=-1\), and between \(x = 0\) and \(x = 1\). For \(f(x)=2x^{3}+9x^{2}+12x + 2\), a zero is between \(x=-1\) and \(x = 0\).