How do you express the function #h(x)=(x + 3)^6# in the form f o g?

Question

HIX Tutor · Accepted Answer

undefined Great question! Let's break it down step by step.

1. **What does "f o g" mean?**
   - It means we have two functions, f and g, and we put them together. We write it as $ f(g(x)) $.

2. **Look at $ h(x) = (x + 3)^6 $**. 
   - What part can we think of as $ g(x) $?

3. **Choose $ g(x) $**:
   - Let’s make $ g(x) = x + 3 $. 
   - This means we are adding 3 to x.

4. **Now, what about $ f(x) $?**
   - Since we want to take $ g(x) $ and put it into another function, we need $ f(x) = x^6 $.
   - This means we will raise whatever comes from $ g(x) $ to the sixth power.

5. **Combine them**:
   - Now we have $ h(x) = f(g(x)) = f(x + 3) = (x + 3)^6 $.

So, can you tell me what functions we have for f and g?

Greyson Brown · Answer

undefined To express the function $ h(x) = (x + 3)^6 $ in the form $ f \circ g $, we need to find two functions $ f(x) $ and $ g(x) $ such that when you apply $ g $ first and then $ f $ to the result, you get $ h(x) $.

Let's choose $ f(x) = x^6 $ and $ g(x) = x + 3 $.

When we apply $ g $ first and then $ f $ to the result, we get:

$$ f(g(x)) = f(x + 3) = (x + 3)^6 $$

So, $ h(x) = (x + 3)^6 $ is expressed in the form $ f \circ g $.

Elijah Allen · Answer

This can be done in an infinite amount of ways. However, the easiest way is to use the sixth power to your advantage. #f@g# means that #g(x)# is being plugged INTO #f(x)#, which means that #f(x)# will have the sixth power in it. If #h(x)=(x+3)^6#, #h(x)=(f@g)(x)# if #f(x)=x^6#
#g(x)=x+3# Work backwards:  to find #(f@g)(x)#, take #g(x)# and put it inside of #f(x)#--you get #(f@g)(x)=(x+3)^6=h(x)#. Other variations include: #f(x)=sqrtx#
#g(x)=(x+3)^12#
#(f@g)(x)=(x+3)^6# #f(x)=x+729#
#g(x)=x^6+18 x^5+135 x^4+540 x^3+1215 x^2+1458 x#
#(f@g)(x)=(x+3)^6#