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4 years ago

Given the function f(x)=x2+2x+1, find: f(2b)

Mathematics

2 answers:

Maru [420]4 years ago

5 0

Hey there!!

Given equation :

f ( x ) = x² + 2x + 1

Find f ( 2b )

In order to solve this question, we will need to replace the value 2b instead of x

... f(2b) = ( 2b )² + 2 ( 2b ) + 1

... f(2b)= 8b² + 4b + 1

Hope my answer helps!

podryga [215]4 years ago

3 0

f(2b)=(2b)2+2(2b)+1

f(2b)=4b+4b+1

f(2b)=8b+1

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3(x-1)=2x+9 what is the equations variable value?

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3 years ago

Find a possible formula for a fourth degree polynomial g that has a double zero at -2, g(4) = 0, g(3) = 0, and g(0) = 12. g(x) =

astraxan [27]

<h2>Answer:</h2>

The possible formula for a fourth degree polynomial g is:

$g(x)=\dfrac{1}{4}(x^4-3x^3-12x^2+20x+48)$

<h2>Step-by-step explanation:</h2>

We know that if a polynomial has zeros as a,b,c and d then the possible polynomial form is given by:

$f(x)=m(x-a)(x-b)(x-c)(x-d)$

Here the polynomial g has a double zero at -2, g(4) = 0, g(3) = 0.

This means that the polynomial g(x) is given by:

$g(x)=m(x-(-2))^2(x-4)(x-3)\\\\i.e.\\\\g(x)=m(x+2)^2(x-4)(x-3)\\\\i.e.\\\\g(x)=m(x^2+2^2+2\times 2\times x)(x(x-3)-4(x-3))\\\\i.e.\\\\g(x)=m(x^2+4+4x)(x^2-3x-4x+12)\\\\i.e.\\\\g(x)=m(x^2+4+4x)(x^2-7x+12)\\\\i.e.\\\\g(x)=m[x^2(x^2-7x+12)+4(x^2-7x+12)+4x(x^2-7x+12)]\\\\i.e.\\\\g(x)=m[x^4-7x^3+12x^2+4x^2-28x+48+4x^3-28x^2+48x]\\\\i.e.\\\\g(x)=m[x^4-7x^3+4x^3+12x^2+4x^2-28x^2-28x+48x+48]\\\\i.e.\\\\g(x)=m[x^4-3x^3-12x^2+20x+48]$

Also,

$g(0)=12$

i.e.

$48m=12\\\\i.e.\\\\m=\dfrac{12}{48}\\\\i.e.\\\\m=\dfrac{1}{4}$

Hence, the polynomial g(x) is given by:

$g(x)=\dfrac{1}{4}(x^4-3x^3-12x^2+20x+48)$

8 0

3 years ago

Hello what is the answer to this and the domain

nlexa [21]

We are given the following functions:

$\begin{gathered} f(x)=7\sqrt[]{x}+6 \\ g(x)=x+6 \end{gathered}$

We are asked to determine the composite function:

$(f\circ g)(x)$

The composition of functions is equivalent to:

$(f\circ g)(x)=f(g(x))$

Therefore, we replace the value of "x" in function "f" for the function "g", therefore, we get:

$(f\circ g)(x)=f(g(x))=7\sqrt[]{x+6}+6$

Since we can't simplify any further this is the composition.

Now we are asked to determine the domain of this function. Since we have a square root, the domain must be the values of "x" where the term inside the radical is greater or equal to zero, therefore, we have: