Question is incomplete
Answer:
See Explanation
Step-by-step explanation:
Let the cubic function be
[tex]f(x) = (x,y)[/tex]
First transformation: Vertical compression
Let the scale factor be a where [tex]0 < |a| < 1[/tex]
The new function becomes:
[tex]f'(x) = (x,ay)[/tex]
Next: Reflection over y-axis
The rule is: [tex](x,y)=>(-x,y)[/tex]
The new function becomes:
[tex]f"(x) = (-x,ay)[/tex]
Lastly: Shifted up
Let the units up be b
The rule is: [tex](x,y)=>(x,y+b)[/tex]
So, the new function becomes:
[tex]f"'(x) = (-x,ay+b)[/tex]
Using the rules states above, assume the cubic function is:
[tex]f(x) = x^3[/tex]
Vertical compress [tex]f(x) = x^3[/tex] by [tex]\frac{1}{2}[/tex]
[tex]f'(x) = \frac{1}{2}x^3[/tex]
Reflect [tex]f'(x) = \frac{1}{2}x^3[/tex] over y-axis
Rule: [tex]f"(x)= f'(-x)[/tex]
Solving f'(-x)
[tex]f'(x) = \frac{1}{2}x^3[/tex]
[tex]f'(-x) = \frac{1}{2}(-x)^3[/tex]
[tex]f'(-x) = -\frac{1}{2}x^3[/tex]
So:
[tex]f"(x) = -\frac{1}{2}x^3[/tex]
Lastly: Shift [tex]f"(x) = -\frac{1}{2}x^3[/tex] by 3
Rule: [tex](x,y)=>(x,y+b)[/tex]
[tex]f'"(x) = f"(x + 3)[/tex]
Solving: f"(x + 3)
[tex]f"(x) = -\frac{1}{2}x^3[/tex]
Substitute x + 3 for x
[tex]f"(x + 3) = -\frac{1}{2}(x + 3)^3[/tex]
So:
[tex]f'"(x) = f"(x + 3)[/tex]
[tex]f"'(x) = -\frac{1}{2}(x + 3)^3[/tex]
