I need help solving this problem f(x)=2x^2+1 find and simplify f(x+2)-f(x)/2?

Answer 1

#f(x+2)-f(x)/2=x^2+8x+17/2#

#f(x) = 2x^2 + 1#
#f(x+2) - f(x)/2#
First, find #f(x+2)# by plugging #x+2# into the original equation
#f(x+2) = 2(x+2)^2 + 1# #f(x+2) = 2(x^2+4x+4) + 1# #f(x+2) = 2x^2+8x+9#
Next, find #f(x)/2# by substituting #2x^2+1# for #f(x)#
#f(x)/2 = (2x^2+1)/2# #f(x)/2 = x^2+1/2#

For the final response, combine your responses via substitution.

#f(x+2) - f(x)/2# #= (2x^2+8x+9) - (x^2+1/2)# #=x^2+8x+17/2#
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Answer 2

#x^2+8x+17/2#

#f(x+2)-f(x)/2# #" "# #=2(x+2)^2+1-(2x^2+1)/2# #" "# #=2(x^2+4x+4)+1-(2x^2)/2-1/2# #" "# #=2x^2+8x+8+1-x^2-1/2# #" "# #=2x^2-x^2+8x+8+1-1/2# #" "# #=x^2+8x+17/2#
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Answer 3

To find and simplify ( \frac{{f(x+2) - f(x)}}{2} ), where ( f(x) = 2x^2 + 1 ):

  1. Substitute ( x+2 ) and ( x ) into ( f(x) ) to get ( f(x+2) ) and ( f(x) ).
  2. Calculate ( f(x+2) - f(x) ).
  3. Divide the result by 2.
  4. Simplify the expression obtained.

Let's proceed:

  1. ( f(x+2) = 2(x+2)^2 + 1 = 2(x^2 + 4x + 4) + 1 = 2x^2 + 8x + 8 + 1 = 2x^2 + 8x + 9 ).

    ( f(x) = 2x^2 + 1 ).

  2. ( f(x+2) - f(x) = (2x^2 + 8x + 9) - (2x^2 + 1) = 8x + 9 - 1 = 8x + 8 ).

  3. ( \frac{{f(x+2) - f(x)}}{2} = \frac{{8x + 8}}{2} = 4x + 4 ).

Therefore, ( \frac{{f(x+2) - f(x)}}{2} = 4x + 4 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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