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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. =)
y=-2x+5 since b is the y intercept so it is 5 and it is a negative slope so 2/-1 is -2
Answer:
-3/4x
Step-by-step explanation:
Down 3, right 4. We know it's negative because it's declining, and going down till you reach the x-coordinate of the point gives us -3, and going right 4 units to the location of the point.
First, let's see if we can rewrite this word problem a little bit more mathematically. We won't get to mathy or technical so no worries. We just want to look at it in a more straightforward way, if we can.
Train A's mph plus Train B's mph summed equal 723.5 mph. Train A's mph is greater than Train B's mph by 12.5 mph.
So what should we do to solve this problem? Since we are dealing with two of something and we know the value of the two combined, it might make sense to start by dividing that value by 2.
723.5 / 2 = <em /> 361.75. If the two trains were travelling at the same speed, we would be done. Unfortunately, they are not so we need to think about this some more.
Train A is going 12.5 mph faster than Train B. Let's rewrite.
Train A mph = 12.5 + 361.75 = 374.25 Okay, so Train A is travelling at a speed of 374.25 mph. So we're done right? Not exactly. We are asked to fing the speeds of BOTH trains. How do we find the speed of Train B? We have added a portion of the combined total to Train A. It seems to follow, then, we should probably subtract the same portion from Train A. What are we going to do? You guessed it! Rewrite.
Train B mph = 361.75 - 12.5 = 349.25 HA HA! We seem to have figured it out. Let's do one last thing to check our work. Let's add the two trains' speeds together. If we did this right, we should get our summed value of 723.5 mph
374.25 + 349.25 = 723.5
Pat yourself on the back! We did it!
374.25 + 349.25 =
Answer:
The statement BI = BK is true from the given information ⇒ B
Step-by-step explanation:
If a line is a perpendicular bisector of a line segment, then
- The line intersects the line segment in 4 right angles
- The line intersects the line segment in the mid-point of the line segment
- Any point on the line is equidistant from the endpoints of the line segment
Let us find the true statement
∵ Line AB is the perpendicular bisector of segment IK
→ By using the 1st note above
∴ AB ⊥ IK
∴ ∠IJA, ∠KJA, ∠IJB, ∠KJB are right angles
→ By using the 2nd note above
∴ J is the mid-point of IK
∴ IJ = JK
∵ Any point on line AB is equidistant from The endpoints of IK ⇒ 3rd note
∴ AI = AK
∴ BI = BK
∴ The statement BI = BK is true from the given information
