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Ket [755]
3 years ago
15

Write a polynomial function of least degree with integral coefficients that has the given zeros. -5, 3i

Mathematics
1 answer:
Virty [35]3 years ago
5 0

Solving for the polynomial function of least degree with integral coefficients whose zeros are -5, 3i

 

We have:
x = -5

Then x + 5 = 0

Therefore one of the factors of the polynomial function is (x + 5)


Also, we have:
x = 3i
Which can be rewritten as:
x = Sqrt(-9)
Square both sides of the equation:
x^2 = -9
x^2 + 9 = 0

Therefore one of the factors of the polynomial function is (x^2 + 9)


The polynomial function has factors: (x + 5)(x^2 + 9)
= x(x^2 + 9) + 5(x^2 + 9)

= x^3 + 9x + 5x^2 = 45

Therefore, x^3 + 5x^2 + 9x – 45 = 0

f(x) = x^3 + 5x^2 + 9x – 45

 

The polynomial function of least degree with integral coefficients that has the given zeros, -5, 3i is f(x) = x^3 + 5x^2 + 9x – 45

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A function f(x)=3x+12.
seraphim [82]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2822258

_______________


•  Function:   f(x) = 3x + 12.


A.  Finding the inverse of f.

The composition of f with its inverse results in the identity function:

(f o g)(x) = x

f[ g(x) ] = x

3 · g(x) + 12 = x

3 · g(x) = x – 12

              x – 12
g(x)  =  ⸺⸺
                 3

               x 
g(x)  =  ⸺  –  4    <———    this is the inverse of f.
               3

________


B.  Verifying that the composition of f and g gives us the identity function:

•  \mathsf{(f\circ g)(x)}

\mathsf{=f\big[g(x)\big]}\\\\\\ \mathsf{=3\cdot \left(\dfrac{x}{3}-4\right)+12}\\\\\\&#10;\mathsf{=\diagup\hspace{-7}3\cdot \dfrac{x}{\diagup\hspace{-7}3}-3\cdot 4+12}\\\\\\&#10;\mathsf{=x-12+12}\\\\&#10;\mathsf{=x\qquad\quad\checkmark}


and also

•  \mathsf{(g\circ f)(x)}

\mathsf{=g\big[f(x)\big]}\\\\\\ \mathsf{=\dfrac{f(x)}{3}-4}\\\\\\ \mathsf{=\dfrac{3x+12}{3}-4}\\\\\\&#10;\mathsf{=\dfrac{\diagup\hspace{-7}3\cdot (x+4)}{\diagup\hspace{-7}3}-4}\\\\\\&#10;\mathsf{=x+4-4}\\\\&#10;\mathsf{=x\qquad\quad\checkmark}

________


C.  Since f and g are inverse, then

f(g(– 2))

= (f o g)(– 2)

= – 2          <span>✔
</span>

•  Call h the compositon of f and g. So,

h(x) = (f o g)(x)

h(x) = x


As you can see above, there is no restriction for h. Therefore, the domain of h is R (all real numbers).


I hope this helps. =)

5 0
3 years ago
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