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2 years ago

Show that 6 – 4x - x^2 can never be greater than 10. ( This is an Additional Math Question )

Mathematics

1 answer:

notsponge [240]2 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

Given

6 - 4x - x²

Since the coefficient of the x² term < 0 then parabola has a maximum vertex.

To find the x- coordinate of the vertex

$x_{vertex}$ = - $\frac{b}{2a}$

with a = - 1 and b = - 4 , then

$x_{vertex}$ = - $\frac{-4}{-2}$ = - 2

The corresponding y- coordinate is found by substituting x = - 2 , that is

6 - 4(- 2) - (- 2)² = 6 + 8 - 4 = 10

vertex = (- 2, 10 )

The maximum value of the function is the y- coordinate of the vertex

Then maximum value is 10

Thus 6 - 4x - x² is never greater than 10

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Use the identity below to complete the tasks:

Gennadij [26K]

Answer:

a3+b3=(a+b)(a2-ab+b2)

4 0

2 years ago

Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

f(x)=cos(x)

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0

2 years ago

Please help me with my math

ValentinkaMS [17]

Im failing math but passing all others im sorry lol

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2 years ago

An airline charges one fee for small luggage bags, and a larger fee for large luggage bags. One family checks five small bags an

RUDIKE [14]

X=small bag
y=large bag

5x+3y=260
4x+2y=200

(5x+3y=260)*4 =20x+12y=1040
(4x+2y=200)*5 =20x+10y=1000

subtract two equations: 2y=40
divide:                           y=20
plug y in                         5x+3*20=260
simplify                         x=40

small bag=$40
large bag=$20

7 0

3 years ago

Read 2 more answers

0.4= Blankx10 please answer quick

Marta_Voda [28]

Answer:

1, 0.4

2, 0.004

3, 0.04

Step-by-step explanation:

hope this helps. :)

6 0

2 years ago