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Answer:
Step-by-step explanation:
In order to write the series using the summation notation, first we need to find the nth term of the sequence formed. The sequence generated by the series is an arithmetic sequence as shown;
4, 8, 12, 16, 20...80
The nth term of an arithmetic sequence is expressed as Tn = a +(n-1)d
a is the first term = 4
d is the common difference = 21-8 = 8-4 = 4
n is the number of terms
On substituting, Tn = 4+(n-1)4
Tn = 4+4n-4
Tn = 4n
The nth term of the series is 4n.
Since the last term is 80, L = 4n
80 = 4n
n = 80/4
n = 20
This shows that the total number of terms in the sequence is 20
According to the series given 4 + 8 + 12 + 16 + 20+ . . . + 80 , we are to take the sum of the first 20terms of the sequence. Using summation notation;
4 + 8 + 12 + 16 + 20+ . . . + 80 =
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_______________
• 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\big[g(x)\big]}\\\\\\ \mathsf{=3\cdot \left(\dfrac{x}{3}-4\right)+12}\\\\\\ \mathsf{=\diagup\hspace{-7}3\cdot \dfrac{x}{\diagup\hspace{-7}3}-3\cdot 4+12}\\\\\\ \mathsf{=x-12+12}\\\\ \mathsf{=x\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7B%3Df%5Cbig%5Bg%28x%29%5Cbig%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D3%5Ccdot%20%5Cleft%28%5Cdfrac%7Bx%7D%7B3%7D-4%5Cright%29%2B12%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3D%5Cdiagup%5Chspace%7B-7%7D3%5Ccdot%20%5Cdfrac%7Bx%7D%7B%5Cdiagup%5Chspace%7B-7%7D3%7D-3%5Ccdot%204%2B12%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx-12%2B12%7D%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%5Cqquad%5Cquad%5Ccheckmark%7D)
and also
•
![\mathsf{=g\big[f(x)\big]}\\\\\\ \mathsf{=\dfrac{f(x)}{3}-4}\\\\\\ \mathsf{=\dfrac{3x+12}{3}-4}\\\\\\ \mathsf{=\dfrac{\diagup\hspace{-7}3\cdot (x+4)}{\diagup\hspace{-7}3}-4}\\\\\\ \mathsf{=x+4-4}\\\\ \mathsf{=x\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7B%3Dg%5Cbig%5Bf%28x%29%5Cbig%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D%5Cdfrac%7Bf%28x%29%7D%7B3%7D-4%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D%5Cdfrac%7B3x%2B12%7D%7B3%7D-4%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3D%5Cdfrac%7B%5Cdiagup%5Chspace%7B-7%7D3%5Ccdot%20%28x%2B4%29%7D%7B%5Cdiagup%5Chspace%7B-7%7D3%7D-4%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%2B4-4%7D%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%5Cqquad%5Cquad%5Ccheckmark%7D)
________
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. =)
_______________
• 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:
•
and also
•
________
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. =)