Question

for the given function, determine consecutive values of x between which each real zero is located. 18. $f(x)=x^{3}-3x + 1$ 19. $f(x)=2x^{3}+9x^{2}+12x + 2$

Explanation:

Step1: Evaluate the function at integer - values

We will evaluate \(f(x)\) at consecutive integer values of \(x\). For \(f(x)=x^{3}-3x + 1\), we start checking values.
When \(x=-2\), \(f(-2)=(-2)^{3}-3\times(-2)+1=-8 + 6+1=-1\).
When \(x=-1\), \(f(-1)=(-1)^{3}-3\times(-1)+1=-1 + 3+1=3\).
When \(x = 0\), \(f(0)=0^{3}-3\times0 + 1=1\).
When \(x = 1\), \(f(1)=1^{3}-3\times1+1=1 - 3+1=-1\).
When \(x = 2\), \(f(2)=2^{3}-3\times2+1=8 - 6+1=3\).

Step2: Use the Intermediate - Value Theorem

The Intermediate - Value Theorem states that if a function \(y = f(x)\) is continuous on a closed interval \([a,b]\) and \(k\) is a number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\) in \((a,b)\) such that \(f(c)=k\). Since \(f(x)\) is a polynomial, it is continuous everywhere.
Since \(f(-2)=-1\) and \(f(-1)=3\), there is a zero between \(x=-2\) and \(x=-1\).
Since \(f(0)=1\) and \(f(1)=-1\), there is a zero between \(x = 0\) and \(x = 1\).
Since \(f(1)=-1\) and \(f(2)=3\), there is a zero between \(x = 1\) and \(x = 2\).

For \(f(x)=2x^{3}+9x^{2}+12x + 2\):
When \(x=-4\), \(f(-4)=2\times(-4)^{3}+9\times(-4)^{2}+12\times(-4)+2=2\times(-64)+9\times16-48 + 2=-128+144-48 + 2=-30\).
When \(x=-3\), \(f(-3)=2\times(-3)^{3}+9\times(-3)^{2}+12\times(-3)+2=2\times(-27)+9\times9-36 + 2=-54 + 81-36 + 2=-7\).
When \(x=-2\), \(f(-2)=2\times(-2)^{3}+9\times(-2)^{2}+12\times(-2)+2=2\times(-8)+9\times4-24 + 2=-16+36-24 + 2=-2\).
When \(x=-1\), \(f(-1)=2\times(-1)^{3}+9\times(-1)^{2}+12\times(-1)+2=-2 + 9-12 + 2=-3\).
When \(x = 0\), \(f(0)=2\times0^{3}+9\times0^{2}+12\times0+2=2\).
Since \(f(-2)=-2\) and \(f(-1)=-3\), no zero here. Since \(f(-1)=-3\) and \(f(0)=2\), there is a zero between \(x=-1\) and \(x = 0\).

Answer:

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