Question

determine whether the intermediate value theorem guarantees that the function has a zero on the given interval.
q(x)=2x^{3}-14x^{2}+7x + 8
(a) (1,2)
(b) (2,3)
(c) (3,4)
(d) (4,5)
part: 0 / 5
(a) (1,2)
to determine if (q(x)=2x^{3}-14x^{2}+7x + 8) has a zero on the interval (1,2) first find (q(1)) and (q(2)).
q(1)=2(1)^{3}-14(1)^{2}+7(1)+8=
q(2)=2(2)^{3}-14(2)^{2}+7(2)+8=
since (q(1)) and (q(2)) select opposite signs, then the intermediate value theorem select the existence of at least one zero on the interval (1,2).

Explanation:

Step1: Calculate \(q(1)\)

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$$

Step2: Calculate \(q(2)\)

$$
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$$

Step3: Apply Intermediate - Value Theorem

Since \(q(1)=3>0\) and \(q(2)= - 18<0\), \(q(1)\) and \(q(2)\) have opposite signs. According to the Intermediate - Value Theorem, if a function \(y = q(x)\) is continuous on the closed interval \([a,b]\) (and a polynomial function is continuous everywhere), and \(q(a)\) and \(q(b)\) have opposite signs, then there is at least one \(c\in(a,b)\) such that \(q(c)=0\).

Step1: Calculate \(q(2)\)

We already know \(q(2)=-18\) from above.

Step2: Calculate \(q(3)\)

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$$

Step3: Apply Intermediate - Value Theorem

Since \(q(2)=-18\) and \(q(3)=-43\) both have negative signs, the Intermediate - Value Theorem does not guarantee the existence of a zero on the interval \([2,3]\).

Step1: Calculate \(q(3)\)

We know \(q(3)=-43\) from above.

Step2: Calculate \(q(4)\)

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$$

Step3: Apply Intermediate - Value Theorem

Since \(q(3)\) and \(q(4)\) both have negative signs, the Intermediate - Value Theorem does not guarantee the existence of a zero on the interval \([3,4]\).

Step1: Calculate \(q(4)\)

We know \(q(4)=-60\) from above.

Step2: Calculate \(q(5)\)

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$$

Step3: Apply Intermediate - Value Theorem

Since \(q(4)\) and \(q(5)\) both have negative signs, the Intermediate - Value Theorem does not guarantee the existence of a zero on the interval \([4,5]\).

Answer:

Since \(q(1)\) and \(q(2)\) have opposite signs, the Intermediate - Value Theorem guarantees the existence of at least one zero on the interval \([1,2]\).

For part (b):