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svet-max [94.6K]
3 years ago
10

Compare and contrast the results of adding or multiplying a value outside the argument of a function to the result of adding or

multiplying a value inside the argument of a function.
Mathematics
1 answer:
ki77a [65]3 years ago
5 0

Answer:

Only in special funcitons, specifically linear functions whose graphs pass through the origin, it is the same to put the constant inside the argument or outside

Step-by-step explanation:

If the function has the form f(x) = a*x, where a is a constant, then we have that f(c*x) = a*(c*x) = c*(a*x) = c*f(x). This kind of functions are proportional to the identity function f(x) = x, and they comprehend the linear functions whose graph pass through the origin.

For other linear function this property isnt true. For example if f(x) = x+4, then

f(4) = 4+4 = 8,

f(2*4) = 8+4 = 12

2*f(4) = 2*8 = 16

Thus, 2*f(4) is not f(2*4).

This property also isnt true for quadratic functions for example. If f(x) = x², then f(1) = 1, thus 3*f(1) = 3, however, f(3*1) = 3² = 9.

There might be coincidences for specific values, for example if f(x) = (x-1)*(x-2), then f(2*1)=2*f(1) = 0, however, for any other constant the result is not the same (for example f(0*1) = (-1)*(-2)=2, and 0*f(1) = 0).

If we want a function to satisfy the property f(c*x) = c*f(x) for any c,x, then it should be true that f(c) = f(c*1) = c*f(1). This means that if f(1) = a, then f(c) = c*a, so, in other words, f(x) = a*x = f(1) * x.

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2(× - 2)=4× - 6(× - 2)​
Bond [772]

Hey there!

Assuming you meant: "\bold{2(x-2=4x-6(x-2)}"

If so follow these steps

  • \bold{Distribute\downarrow}
  • \bold{2(x)+2(-2)=4(x)+(-6)x+(-6))(-2)}
  • \bold{2(x)=2x}
  • \bold{2(-2)=-4}
  • \bold{4x=4x}
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Good luck on your assignment and enjoy your day

~\frak{LoveYourselfFirst:)}

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