Is given a function:
f(x)=x2 + px + q
Prove that:
f(x+1) + f(x-1)= 2+2 f(x)
just plug in the values:
f(x+1) = (x+1)^2 + p(x+1) + q
f(x-1) = (x-1)^2 + p(x-1) + q
add them to get
2x^2 + 2 + 2px + 2q
2(x^2+px+q) + 2
2f(x) + 2
Steve thanks very much, may i have your email id so i can ask a question with integrals/limits which i cant write here
To prove the given equation:
Step 1: Start with the left side of the equation: f(x+1) + f(x-1).
Step 2: Substitute the given function f(x) = x^2 + px + q into the left side of the equation.
So, f(x+1) + f(x-1) = (x+1)^2 + p(x+1) + q + (x-1)^2 + p(x-1) + q.
Step 3: Simplify the equation.
Expanding the squares:
= (x^2 + 2x + 1) + (px + p) + q + (x^2 - 2x + 1) + (px - p) + q.
= 2x^2 + 2px + 2 + 2q.
Step 4: Now, let's simplify the right side of the equation: 2 + 2f(x).
Substituting the given function f(x) = x^2 + px + q:
= 2 + 2(x^2 + px + q)
= 2 + 2x^2 + 2px + 2q.
Step 5: Simplifying the equation:
2x^2 + 2px + 2 + 2q.
Step 6: Compare the left and right side of the equation. We can see that the left side (2x^2 + 2px + 2 + 2q) is equal to the right side (2x^2 + 2px + 2 + 2q).
Therefore, we have proved that f(x+1) + f(x-1) = 2 + 2f(x) for the given function f(x) = x^2 + px + q.