Question

use the intermediate value theorem to check whether the polynomial function p(x) given below has a real zero in the interval 1, 2. if it has a zero in the interval, approximate the zero to two decimal places.
p(x)=2x^5 - x^2 - 14
select the correct choice below and fill in the answer box with your choice if necessary.
a. the polynomial function p(x) has no zero in the given interval.
b. the approximate value of zero of p(x) is
(round to two decimal places as needed.)

Explanation:

Step1: Evaluate \(P(x)\) at \(x = 1\)

Substitute \(x = 1\) into \(P(x)=2x^{5}-x^{2}-14\).
\[ \begin{align*} P(1)&=2\times(1)^{5}-(1)^{2}-14\\ &=2 - 1-14\\ &=- 13 \end{align*} \]

Step2: Evaluate \(P(x)\) at \(x = 2\)

Substitute \(x = 2\) into \(P(x)=2x^{5}-x^{2}-14\).
\[ \begin{align*} P(2)&=2\times(2)^{5}-(2)^{2}-14\\ &=2\times32-4 - 14\\ &=64-4-14\\ &=46 \end{align*} \]
Since \(P(1)=-13<0\) and \(P(2) = 46>0\), by the Intermediate Value Theorem, \(P(x)\) has a zero in the interval \([1,2]\).

Step3: Use the bisection - method (initial step)

Let \(a = 1\), \(b = 2\), and \(c=\frac{a + b}{2}=\frac{1 + 2}{2}=1.5\)
Evaluate \(P(c)\) where \(c = 1.5\)
\[ \begin{align*} P(1.5)&=2\times(1.5)^{5}-(1.5)^{2}-14\\ &=2\times7.59375-2.25-14\\ &=15.1875-2.25 - 14\\ &=-1.0625 \end{align*} \]
Since \(P(1.5)=-1.0625<0\), the zero is in the interval \([1.5,2]\)

Step4: Next iteration of the bisection - method

Let \(a = 1.5\), \(b = 2\), \(c=\frac{1.5 + 2}{2}=1.75\)
Evaluate \(P(1.75)\)
\[ \begin{align*} P(1.75)&=2\times(1.75)^{5}-(1.75)^{2}-14\\ &=2\times16.41308594-3.0625-14\\ &=32.82617188-3.0625 - 14\\ &=15.76367188 \end{align*} \]
Since \(P(1.75)>0\), the zero is in the interval \([1.5,1.75]\)

Step5: Next iteration of the bisection - method

Let \(a = 1.5\), \(b = 1.75\), \(c=\frac{1.5 + 1.75}{2}=1.625\)
Evaluate \(P(1.625)\)
\[ \begin{align*} P(1.625)&=2\times(1.625)^{5}-(1.625)^{2}-14\\ &=2\times11.07006836-2.640625-14\\ &=22.14013672-2.640625 - 14\\ &=5.49951172 \end{align*} \]
Since \(P(1.625)>0\), the zero is in the interval \([1.5,1.625]\)

Step6: Next iteration of the bisection - method

Let \(a = 1.5\), \(b = 1.625\), \(c=\frac{1.5 + 1.625}{2}=1.5625\)
Evaluate \(P(1.5625)\)
\[ \begin{align*} P(1.5625)&=2\times(1.5625)^{5}-(1.5625)^{2}-14\\ &=2\times7.59375\times(1.5625)^{2}-2.44140625-14\\ &=2\times7.59375\times2.44140625-2.44140625-14\\ &=37.03613281-2.44140625 - 14\\ &=20.59472656 \end{align*} \]
Since \(P(1.5625)>0\), the zero is in the interval \([1.5,1.5625]\)

Step7: Next iteration of the bisection - method

Let \(a = 1.5\), \(b = 1.5625\), \(c=\frac{1.5 + 1.5625}{2}=1.53125\)
Evaluate \(P(1.53125)\)
\[ \begin{align*} P(1.53125)&=2\times(1.53125)^{5}-(1.53125)^{2}-14\\ &=2\times(1.53125)^{2}\times(1.53125)^{3}-2.3447265625-14\\ &=2\times2.3447265625\times3.5998535156-2.3447265625-14\\ &=16.91906738-2.3447265625 - 14\\ &=0.5743408175 \end{align*} \]
Since \(P(1.53125)>0\), the zero is in the interval \([1.5,1.53125]\)

Step8: Next iteration of the bisection - method

Let \(a = 1.5\), \(b = 1.53125\), \(c=\frac{1.5 + 1.53125}{2}=1.515625\)
Evaluate \(P(1.515625)\)
\[ \begin{align*} P(1.515625)&=2\times(1.515625)^{5}-(1.515625)^{2}-14\\ &=2\times(1.515625)^{2}\times(1.515625)^{3}-2.29748535156-14\\ &=2\times2.29748535156\times3.4189453125-2.29748535156-14\\ &=15.69777527-2.29748535156 - 14\\ &=-0.6 \end{align*} \]
Since \(P(1.515625)<0\), the zero is in the interval \([1.515625,1.53125]\)

Step9: Next iteration of the bisection - method

Let \(a = 1.515625\), \(b = 1.53125\), \(c=\frac{1.515625 + 1.53125}{2}=1.5234375\)
Evaluate \(P(1.5234375)\)
\[ \begin{align*} P(1.5234375)&=2\times(1.5234375)^{5}-(1.5234375)^{2}-14\\ &=2\times(1.5234375)^{2}\times(1.5234375)^{3}-2.32186096191-14\\ &=2\times2.32186096191\times3.47900390625-2.32186096191-14\\ &=16.12773716-2.32186096191 - 14\\ &=-0.19412380191 \end{align*} \]
Since \(P(1.5234375)<0\), the zero is in the interval \([1.5234375,1.53125]\)

Step10: Next iteration of the bisection - method

Let \(a = 1.5234375\), \(b = 1.53125\), \(c=\frac{1.5234375+1.53125}{2}=1.52734375\)
Evaluate \(P(1.52734375)\)
\[ \begin{align*} P(1.52734375)&=2\times(1.52734375)^{5}-(1.52734375)^{2}-14\\ &=2\times(1.52734375)^{2}\times(1.52734375)^{3}-2.33277587891-14\\ &=2\times2.33277587891\times3.51397705078-2.33277587891-14\\ &=16.38933325-2.33277587891 - 14\\ &=0.05655737109 \end{align*} \]
Since \(P(1.52734375)>0\), the zero is in the interval \([1.5234375,1.52734375]\)

Answer:

B. The approximate value of zero of \(P(x)\) is \(1.52\) (rounded to two decimal places)