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    Given
    f(x)= e^2x +3
    g(x)= ln(x-1)

    Find fg and state it's range. I got the to fg being:
    (x+1)^2 +3 which is what the answers says
    However, I expanded the brackets to get x^2-2x+4. I got the range as fg is greater than or equal to 4 because that's the minimum point of the graph but the answer is fg is greater than 3. Can someone explain why this is?
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    (Original post by Year11guy)
    Given
    f(x)= e^2x +3
    g(x)= ln(x-1)

    Find fg and state it's range. I got the to fg being:
    (x+1)^2 +3 which is what the answers says
    However, I expanded the brackets to get x^2-2x+4. I got the range as fg is greater than or equal to 4 because that's the minimum point of the graph but the answer is fg is greater than 3. Can someone explain why this is?
    Just because the y intercept is 4 doesn't mean that (0,4) is the minimum point.

    Try using your differentiation or complete the square to find the minimum point.
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    (Original post by The_Big_E)
    Just because the y intercept is 4 doesn't mean that (0,4) is the minimum point.

    Try using your differentiation or complete the square to find the minimum point.
    It's okay I figured it out. Thanks for replying anyway.
 
 
 
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Updated: October 23, 2015
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