The value of\[\left( 1-{1\over3} \right)\left( 1-{1\over4} \right)\left( 1-{1\over5} \right)....\left( 1-{1\over n} \right)\]is
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1)
g(x) = f(x)+12 represents translating f(x) 12 units up to get g(x). This is because y = f(x), so we're effectively adding 12 to each y coordinate of the points on f(x) to have them move to the points on g(x).
note: the term "translating" is another way of saying "shifting" or "sliding".
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2)
h(x) = f(x)-7 means translating f(x) 7 units down to get h(x). We use the same reasoning as in problem 1. The general form is h(x) = f(x)+k, and in this case, k = -7. The negative k value tells us to shift down.
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3)
g(x) = f(x+8) means we translate 8 units to the left. The general template is g(x) = f(x-h). If h > 0, then we shift right. If h < 0, then we shift left. The amount that is shifted is equal to the absolute value of h.
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4)
h(x) = f(x-14) tells us to shift f(x) 14 units to the right. Same idea as problem 3, but now h = 14.
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5)
g(x) = 4*f(x) means we vertically stretch f(x) by a factor of 4. Since y = f(x), we are effectively doing g(x) = 4*f(x) = 4*y, which pulls the y coordinates upward by a factor of 4. A point like (1,2) will jump to (1,8) after multiplying the y coordinate by 4.
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6)
g(x) = f(5x) tells us to horizontally compress f(x) by a factor of 5. The general template is g(x) = f(a*x) where the 'a' determines horizontal stretching or compression. If , then we have horizontal stretching going on. If
, then we'll have horizontal compression.