Let p(x) be a polynomial of degree 4 having extremum at x = 1 ,2 and
[tex]\displaystyle\sf \lim_{x\to 0}\left( 1+\dfrac{p(x)}{x^2}\right) = 2 [/tex]
Then the value of p(2) is ?
Answers
PLEASE CHECK THE ATTACHED FILE
![Let P(x) Be A Polynomial Of Degree 4 Having Extremum At X = 1 ,2 And [tex]\displaystyle\sf \lim_{x\to](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/aOScFm3g62FWugWCh2ZTWPPajXO8iNXL.jpeg)
![Let P(x) Be A Polynomial Of Degree 4 Having Extremum At X = 1 ,2 And [tex]\displaystyle\sf \lim_{x\to](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/Kpych4HzOqmDVS53SezYkmklvpKqNinj.jpeg)
We are given that ;
[tex]{\quad \longrightarrow \displaystyle \sf \lim_{x\to 0}\bigg\{1+\dfrac{p(x)}{x^2}\bigg\}=2}[/tex]
Where p(x) is a polynomial of degree 4 , it will help us later, but let's do some manipulations first ;
Can be further written as ;
[tex]{:\implies \quad \displaystyle \sf 1+\lim_{x\to 0}\bigg\{\dfrac{p(x)}{x^2}\bigg\}=2}[/tex]
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 0}\bigg\{\dfrac{p(x)}{x^2}\bigg\}=1}[/tex]
So, here p(x) is polynomial of degree 4, so it will be a biquadratic polynomial, so we will write p(x) in the form of general biquadratic polynomial, so p(x) = ax⁴ + bx³ + cx² + dx + e
Now, first find p(x)/x² ;
[tex]{:\implies \quad \sf \dfrac{p(x)}{x^2}=\dfrac{ax^{4}+bx^{3}+cx^{2}+dx+e}{x^2}}[/tex]
[tex]{:\implies \quad \sf \dfrac{p(x)}{x^2}=ax^{2}+bx+c+\dfrac{d}{x}+\dfrac{e}{x^2}}[/tex]
So now, as [tex]{\bf{x\to 0}}[/tex] , for any values of a, b, c, d and e, the RHS will approach ∞ iff d ≠ e ≠ 0, as the denominator of d and e will be approaching 0 and so the whole limit will be ∞ , but we want the limit to be approaching 1, so when if d = e = 0, the denominator of d and e will be approaching 0 (not absolutely 0), and if d = e = 0, we will have the limit be approaching ax²+ bx + c for x approaching 0 being the limit 1 , and for any values of a, b and c . So now we have ;
[tex]{:\implies \quad \displaystyle \bf \lim_{x\to 0}\dfrac{p(x)}{x^{2}}=\begin{cases}\bf \infty \:, iff\: d\neq e\neq 0 \\ \\ \bf 1\:, iff\: d=e=0\end{cases}}[/tex]
So, now we had to consider the second case, for which the limit is approaching 1, for d = e = 0, so the limand here will just be ax² + bx + c
Now, we so have ;
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 0}ax^{2}+bx+c=1}[/tex]
Putting the limit we will have ;
[tex]{:\implies \quad \sf c=1}[/tex]
So, now as p(x) have extremum at 1 and 2, so p'(x) = 0, for x = 1, 2 , so now finding p'(x)
[tex]{:\implies \quad \sf p(x)=ax^{4}+bx^{3}+x^{2}\quad \qquad \{\because c=1\: and\: d=e=0\}}[/tex]
So, differentiating both sides wr.t.x ;
[tex]{:\implies \quad \sf p^{\prime}(x)=4ax^{3}+3bx^{2}+2x}[/tex]
Now, p'(1) and p'(2) must be 0
[tex]{:\implies \quad \sf p^{\prime}(1)=4a+3b+2=0}[/tex]
[tex]{:\implies \quad \sf p^{\prime}(2)=32a+12b+4=0}[/tex]
So, now we have ;
[tex]{\quad \longrightarrow \displaystyle \begin{cases}\bf 4a+3b=-2 \\ \\ \bf 32a+12b=-4\end{cases}}[/tex]
On multiplying first equation by 8 on both sides we can thus obtain ;
[tex]{\quad \longrightarrow \displaystyle \begin{cases}\bf 32a+24b=-16 \\ \\ \bf 32a+12b=-4\end{cases}}[/tex]
On solving both the equations we will be having ;
[tex]{\quad \longrightarrow \displaystyle \begin{cases}\bf a=\dfrac{1}{4} \\ \\ \bf b=-1\end{cases}}[/tex]
So , now as d = e = 0, c = 1, a = (1/4), b = -1, so putting all the values in p(x) we can obtain p(x) as ;
[tex]{:\implies \quad \sf p(x)=\dfrac{1}{4}(x)^{4}-(x)^{3}+x^{2}}[/tex]
Now, at x = 2 ;
[tex]{:\implies \quad \sf p(2)=\dfrac{1}{4}(2)^{4}-(2)^{3}+2^{2}}[/tex]
[tex]{:\implies \quad \sf p(2)=4-8+4=8-8}[/tex]
[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{p(2)=0}}}[/tex]
This is the required answer
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Important information:
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a
Step-by-step explanation:
Answer:
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Step-by-step explanation:
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Step-by-step explanation:
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Step-by-step explanation
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Answer:
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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Answer:
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Answer:
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Hi friend! Hope you find this response helpful! :)
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Step-by-step explanation:
Answer:
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Answer:
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Check the picture below.
The polygons are similar. Find the value of x.
2-
Р
X
X
20
21
14
5
Q 12 R
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For the given similar polygons, the value of x equal to 30.
Similar PolygonsTwo polygons are classified as similar when their angles are congruent and their sides are proportional.
If the polygons of this exercise are similar. You can write:
[tex]\frac{LK}{QR}=\frac{JK}{PR}=\frac{JL}{PQ}[/tex]
Then,
[tex]\frac{LK}{QR}=\frac{JK}{PR}=\frac{JL}{PQ}\\ \\ \frac{18}{12}=\frac{x}{20}=\frac{21}{14}[/tex]
It is possible to note that the scale factor is [tex]\frac{3}{2}[/tex]. See it.
[tex]\frac{18:6}{12:6}=\frac{3}{2} \\ \\ \frac{21:7}{14:7}=\frac{3}{2}[/tex]
Therefore, you can find x from:
[tex]\frac{18}{12}=\frac{x}{20}\\ \\ 12x=18*20\\ \\ 12x=360\\ \\ x=\frac{360}{12} =30[/tex]
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Answer:
$1.80
Step-by-step explanation:
9x0.2=1.80
He should leave 1.80 as a tip.
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Answer:
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Answers
Solution:
We know that:
Perimeter of rectangle = 2L + 2BGiven:
L = 9 cmW = 5 cmSubstitute the values into the expression to find the perimeter.
Perimeter of rectangle = 2L + 2B=> 2(9) + 2(5)=> 18 + 10=> 28 cmAnswer: 28cm
Step-by-step explanation:
The perimeter formula for a rectangle = 2(l + w); where:
length = 9cm
width = 5cm
P = 2(l + w)
= 2(9 + 5)
= 18 + 10
= 28cm
Therefore, the perimeter of the cheese is 28cm
Find the distance between the two points rounding to the nearest tenth
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Answer:
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Step-by-step explanation:
The first terms of a geometric sequence are as follows.
-810,-270,-90
Find the next two terms of this sequence.
Give exact values (not decimal approximations)
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The fourth and fifth term of this sequence is -30 and -10 respectively.
Data;
first term (a) = -810common difference (r) = ?Common DifferenceThe common difference in a geometric progression is the ratio between two successive terms.
Let's find the common difference in this sequence
[tex]r = \frac{-270}{-810} \\r = \frac{1}{3}[/tex]
The nth term of a geometric sequence is given as
[tex]T_n = ar^n^-^1\\[/tex]
The next two terms of this sequence will be 4th and 5th term.
The fourth term of this sequence is
[tex]T_4 = ar^4^-^1\\T_4 = -810 * (\frac{1}{3})^3\\ T_4 = -810 * \frac{1}{27}\\ T_4 = -30[/tex]
The fifth term of this sequence is
[tex]T_5 = ar^5^-^1\\T_5 = -810 * (\frac{1}{3})^4\\T_5 = -810 * \frac{1}{81} \\T_5 = -10[/tex]
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Which expression(s) below are equivalent to -3x + 6? select all that apply.
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c. -3(x - 2)
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Answers
Answer:
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Answers
Answer:
r = 0.84
Step-by-step explanation:
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Answers
Answer:
9
Step-by-step explanation:
81= 1,3,9,27,81
72= 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36,72
9 is the most common so GCF will be 9.
Answer:
See below.
Step-by-step explanation:
81 - 1,3,9,27,81
72 - 1,2,3,4,6,8,9,12,18,24,36,72
The GCF is 9.
-hope it helps
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Answers
Answer:
ask your teacher for help , maybe youll find the answer quickly
Step-by-step explanation: