Nora went shopping for a new phone because of a sale. The store was offering a 10% discount. If the price on the tag was $24, wh
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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. =)
We can see that revolving the region formed by intersecting 3 lines, we will get 2 cones that are connected their bases.
Volume of the cone V=1/3 *πr²*h
1) small cone has r=5, and h=5
Volume small cone V1= 1/3 *π*5²*5 = 5³/3 *π
2) large cone has r=5, and h=21-6=15, h=15
Volume large cone V2= 1/3 *π*5²*15 = 5³*π
3) whole volume
5³/3 *π + 5³*π=5³π(1/3+1)=((5³*4)/3)π=(500/3)π≈166.7π≈523.6
Area
we see 2 right triangles,
Area of the triangle=1/2*b*h, where b -base, h -height
1) small one, b=5, h=5
A1=(1/2)*5*5=25/2
2)large one, b=5, h=15
A2=(1/2)*5*15=75/2
3) whole area=A1+A2=25/2+75/2=100/2=50
Volume of the cone V=1/3 *πr²*h
1) small cone has r=5, and h=5
Volume small cone V1= 1/3 *π*5²*5 = 5³/3 *π
2) large cone has r=5, and h=21-6=15, h=15
Volume large cone V2= 1/3 *π*5²*15 = 5³*π
3) whole volume
5³/3 *π + 5³*π=5³π(1/3+1)=((5³*4)/3)π=(500/3)π≈166.7π≈523.6
Area
we see 2 right triangles,
Area of the triangle=1/2*b*h, where b -base, h -height
1) small one, b=5, h=5
A1=(1/2)*5*5=25/2
2)large one, b=5, h=15
A2=(1/2)*5*15=75/2
3) whole area=A1+A2=25/2+75/2=100/2=50