Question 94843
Given:
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{{{f(x) = 2 - x^2}}} <=== call this equation 1
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Next you need to find {{{f(x+h)}}}. What this means is you return to {{{f(x) = 2 - x^2}}} and
you replace every x in that equation with {{{x+h}}}. When you do that replacement you get:
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{{{f(x+h) = 2 - (x+h)^2}}} <=== call this equation 2
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To simplify this you need to square the {{{x+h}}} on the right side. Do this by multiplying:
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{{{(x+h)^2 = (x+h)*(x+h) = x^2 +2xh + h^2}}}
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Substitute {{{x^2 +2xh + h^2}}} into equation 2 in place of {{{(x+h)^2}}} and equation 2 becomes:
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{{{f(x+h) = 2 - (x^2 + 2xh + h^2) = 2 - x^2 - 2xh - h^2}}} <=== call this equation 3
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The problem asked you to find {{{f(x+h) - f(x)}}}. From equation 3 we know that:
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{{{f(x+h) = 2 - x^2 - 2xh - h^2}}}
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and from equation 1 we know:
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{{{f(x) = 2 - x^2}}}
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So we substitute the right side of these two equations into {{{f(x+h) - f(x)}}} in place
of {{{f(x+h)}}} and {{{f(x)}}} to get:
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{{{f(x+h) - f(x) = 2 - x^2 - 2xh - h^2 - (2 - x^2)}}}
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On the right side, since the set of parentheses are preceded by a minus sign, the parentheses
can be removed if we change the signs of the two terms in the parentheses. This makes the
equation become:
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{{{f(x+h) - f(x) = 2 - x^2 - 2xh - h^2 - 2 + x^2)}}}
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Now on the right side you can notice that the +2 and the -2 cancel each other as does the 
{{{-x^2 }}} and the {{{x^2}}}. Those terms, therefore, disappear and you are left with:
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{{{f(x+h) - f(x) = - 2xh - h^2}}}
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Hope this helps you to understand the problem a little more.
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