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Thread: Functions

  1. #1
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    Functions

    Let the function f : [3,1) -> R be defined by f(z) = (z −3)^2 +5, and
    g : R -> R be defined by g(x) = 7 e^x.
    (a) Find ran f.
    (b) Find g(ran f).
    (c) Let B := ran f and C := g(ran f). Write explicitly the expression for (g)^−1(h) for h E C,
    and the expression for (f)^−1(x) for x E B.
    (d) Write explicitly the expression for (g o f)^−1(h) for h E C.
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  2. #2
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    Quote Originally Posted by luckyc1423 View Post
    Let the function f : [3,1) -> R be defined by f(z) = (z −3)^2 +5, and
    g : R -> R be defined by g(x) = 7 e^x.
    (a) Find ran f.
    What is $\displaystyle [3,1)=\{\}$?
    I assume $\displaystyle [1,3)$.

    The range are all $\displaystyle y\in \mathbb{R}$ such that $\displaystyle f(z)=y$ for some $\displaystyle z\in [1,3)$.
    Thus,
    $\displaystyle f(z)=y$
    $\displaystyle (z-3)^2+5=y$
    $\displaystyle (z-3)^2=y-5$
    To have a solution we need that, $\displaystyle y-5\geq 0$.
    Thus,
    $\displaystyle z=3\pm \sqrt{y-5}$
    We need that,
    $\displaystyle 1\leq 3\pm \sqrt{y-5} <3$
    $\displaystyle -2\leq \pm \sqrt{y-5}<0$
    We see that $\displaystyle +$ cannot work.
    Thus,
    $\displaystyle -2\leq -\sqrt{y-5}<0$ this tells us that $\displaystyle y\not = 5$.
    It is equivalnet to squaring both sides,
    $\displaystyle 4\geq y-5<0$
    $\displaystyle 9\geq y < 5$
    Note it satisfies the condition we have before $\displaystyle y-5\geq 0$.
    (b) Find g(ran f).
    This is defined as,
    $\displaystyle g: (5,9]\to \mathbb{R}$
    $\displaystyle g(x)=7e^x$
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