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 $q(2)$.

$q(1)=2(1)^{3}-14(1)^{2}+7(1)+8=square$

$q(2)=2(2)^{3}-14(2)^{2}+7(2)+8=square$

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)

$q(1)=2\times(1)^3 - 14\times(1)^2+7\times(1)+8=2 - 14 + 7+8=3$

Step2: Calculate q(2)

$q(2)=2\times(2)^3 - 14\times(2)^2+7\times(2)+8=2\times8-14\times4 + 14+8=16-56 + 14+8=-18$

Step3: Check signs

Since $q(1) = 3>0$ and $q(2)=-18<0$, they have opposite signs.

Answer:

(a) The intermediate - value theorem guarantees the existence of at least one zero on the interval $[1,2]$.
(b)

Step1: Calculate q(2) and q(3)

We already know $q(2)=-18$.
$q(3)=2\times(3)^3 - 14\times(3)^2+7\times(3)+8=2\times27-14\times9 + 21+8=54-126 + 21+8=-43$
Since $q(2)=-18$ and $q(3)=-43$ have the same sign, the intermediate - value theorem does not guarantee a zero on the interval $[2,3]$.
(c)

Step1: Calculate q(3) and q(4)

We know $q(3)=-43$.
$q(4)=2\times(4)^3 - 14\times(4)^2+7\times(4)+8=2\times64-14\times16 + 28+8=128-224 + 28+8=-60$
Since $q(3)=-43$ and $q(4)=-60$ have the same sign, the intermediate - value theorem does not guarantee a zero on the interval $[3,4]$.
(d)

Step1: Calculate q(4) and q(5)

We know $q(4)=-60$.
$q(5)=2\times(5)^3 - 14\times(5)^2+7\times(5)+8=2\times125-14\times25 + 35+8=250-350 + 35+8=-57$
Since $q(4)=-60$ and $q(5)=-57$ have the same sign, the intermediate - value theorem does not guarantee a zero on the interval $[4,5]$.