determine whether the intermediate value theorem guarantees that the function has a zero on the given…

determine whether the intermediate value theorem guarantees that the function has a zero on the given interval.\nq(x)=2x^{3}-14x^{2}+7x + 8\n(a) (1,2)\n(b) (2,3)\n(c) (3,4)\n(d) (4,5)\npart: 0 / 5\n(a) (1,2)\nto 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)).\nq(1)=2(1)^{3}-14(1)^{2}+7(1)+8=\nq(2)=2(2)^{3}-14(2)^{2}+7(2)+8=\nsince (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).

determine whether the intermediate value theorem guarantees that the function has a zero on the given interval.\nq(x)=2x^{3}-14x^{2}+7x + 8\n(a) (1,2)\n(b) (2,3)\n(c) (3,4)\n(d) (4,5)\npart: 0 / 5\n(a) (1,2)\nto 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)).\nq(1)=2(1)^{3}-14(1)^{2}+7(1)+8=\nq(2)=2(2)^{3}-14(2)^{2}+7(2)+8=\nsince (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).

Answer

Explanation:

Step1: Calculate (q(1))

[ \begin{align*} q(1)&=2(1)^3 - 14(1)^2+7(1)+8\ &=2 - 14 + 7+8\ &=(2 + 7+8)-14\ &=17 - 14\ &=3 \end{align*} ]

Step2: Calculate (q(2))

[ \begin{align*} q(2)&=2(2)^3-14(2)^2 + 7(2)+8\ &=2\times8-14\times4 + 14+8\ &=16-56+14 + 8\ &=(16+14+8)-56\ &=38 - 56\ &=-18 \end{align*} ]

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).

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):

Explanation:

Step1: Calculate (q(2))

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

Step2: Calculate (q(3))

[ \begin{align*} q(3)&=2(3)^3-14(3)^2 + 7(3)+8\ &=2\times27-14\times9+21 + 8\ &=54-126+21 + 8\ &=(54+21+8)-126\ &=83 - 126\ &=-43 \end{align*} ]

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]).

Answer:

Since (q(2)) and (q(3)) do not have opposite signs, the Intermediate - Value Theorem does not guarantee the existence of a zero on the interval ([2,3]).

For part (c):

Explanation:

Step1: Calculate (q(3))

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

Step2: Calculate (q(4))

[ \begin{align*} q(4)&=2(4)^3-14(4)^2 + 7(4)+8\ &=2\times64-14\times16+28 + 8\ &=128-224+28 + 8\ &=(128+28+8)-224\ &=164 - 224\ &=-60 \end{align*} ]

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]).

Answer:

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

For part (d):

Explanation:

Step1: Calculate (q(4))

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

Step2: Calculate (q(5))

[ \begin{align*} q(5)&=2(5)^3-14(5)^2 + 7(5)+8\ &=2\times125-14\times25+35 + 8\ &=250-350+35 + 8\ &=(250+35+8)-350\ &=293 - 350\ &=-57 \end{align*} ]

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