f(x)=2x^2+3 find and simplify f(x+h)-f(x)
\[f(x) = 2x^2+3\]\[f(x+h) = 2(x+h)^2 + 3\]\[f(x+h) - f(x) = 2(x+h)^2 + 3 - (2x^2 +3) = 2(x+h)^2 + 3 - 2x^2 -3\] Expand the \((x+h)^2\) bit and collect like terms.
I got 2(x+h)^2-2x^2
You have to multiply out the (x+h)^2 part...
Trust me, you're going to see much yuckier stuff doing derivatives with the definition than this :-)
I need help with that part I keep getting confused with my math
\[(x+h)^2 = (x+h)(x+h)\] right? You can multiply that out, can't you?
x^2+hx+h?
right
\[(x+h)(x+h) = x^2 + hx + hx + h^2 = x^2 + 2hx + h^2\]
so then I got 2+x^2+2hx+h^2-2x^2
Remember, each term of the first one multiplies each term of the second one. Some people like the FOIL acronym: First Outer Inner Last I prefer to think of it as \[(x+h)(x+h) = x(x+h) + h(x+h) = x^2 + hx + hx + h^2\]Then I don't have to think any differently if I'm doing something like \((x + y + z)(x + y + z)\)
ok is my previous post correct?
\[f(x+h) - f(x) = 2(x+h)^2 + 3 - 2x^2 -3 = 2(x^2+2hx+h^2) -2x^2 = \]\[2x^2 - 2x^2+4hx+2h^2 = 4hx+2h^2\]
is that the answer
\(4hx+2h^2\) is the final result, yes.
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