simplify each expression if f(x)=x2-1f(x-3)

Question

I have to find and simplify each expression if f(x)=x² -1

f(x-3)

f(4+h) - f(4)

Answer 1

Note that the function factors by the difference of two squares: f(x) =  (x+1)(x-1).

Thus:

f(x-3) = (x-2)(x-4) = x²-6x+8 and

f(4+h) - f(4) = (5+h)(3+h) - (5)(3) = h²+8h

 

Answer 2

Hi, Kristin.

f(x) = x² - 1 is a function, which just tells us what to do with x.

Whatever is inside the parentheses gets substituted for x in the equation. For example:

f(3) = 3² - 1   so   f(3) = 8

Then   f(x-3)  means that we substitute  x-3   in for x, and simplify:

f(x-3) = (x - 3)² -1

We can multiply (x-3)(x-3) and get:

f(x-3) = (x² -6x + 9) - 1     which simplifies to:

f(x-3) = x² - 6x + 8

 

The second expression has two parts, and I will do the f(4+h) first:

f(4+h) = (4 + h)² -1

Mutiplying and simplifying gives us:    f(4+h) = h² + 8h + 15

f(4) = 4² - 1   -->   f(4) = 15

So combining them:

f(4+h) - f(4) = h² + 8h + 15 -15

f(4+h) - f(4) = h² + 8h     which factors to:

f(4+h) - f(4) = h(h + 8)

Hope this helps!

 

Answer 3

when a function ƒ is defined in terms of x (the input), it is expressed as ƒ(x)....

for ƒ(x-3), x-3 is the new input of the function ƒ...

so if ƒ(x)= x²-1, then ƒ(x-3)= (x-3)²-1 and you solve from there

and for ƒ(4+h) - ƒ(4), you can solve for each ƒ separately then subtract:

  if ƒ(x)= x²-1,           then ƒ(4+h)= (4+h)²-1     and       ƒ(4)= (4)²-1

  so,   ƒ(4+h) - ƒ(4) = ((4+h)²-1) - ((4)²-1) = (4+h)² - 16