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3 years ago

Solve y=1/4x-1 if the domain is {-4,-2,0,2,4}.

Mathematics

1 answer:

kipiarov [429]3 years ago

5 0

Allowed values for y: $-2,-\frac{3}{2},-1,-\frac{1}{2},0$

Step-by-step explanation:

The domain of a function $f(x)$ represents the set of values for which the function is defined. In other words, it represents the values of x that are allowed.

The function in this problem is:

$y=\frac{1}{4}x-1$

And the domain is

D: {-4,-2,0,2,4}

This means that these are the only allowed values for x. We can find the range of the functions (the possible values for y) simply by substituting these values of x into the function. We get:

For x = -4,

$y=\frac{1}{4}(-4)-1=-1-1=-2$

For x = -2,

$y=\frac{1}{4}(-2)-1=-\frac{1}{2}-1=-\frac{3}{2}$

For x = 0,

$y=\frac{1}{4}(0)-1=-1$

For x = 2,

$y=\frac{1}{4}(2)-1=\frac{1}{2}-1=-\frac{1}{2}$

For x = 4,

$y=\frac{1}{4}(4)-1=1-1=0$

Therefore, the allowed values for y are:

$-2,-\frac{3}{2},-1,-\frac{1}{2},0$

Learn more about domain and range:

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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

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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

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