Grpahing functions

**realintegerz** · October 13th 2008, 02:35 PM

I'm having a hard time understanding these

Let g(x) = 3(x+2)^2 + 4
a. Graph y=gx without a calculator (how do I graph it???, we haven't learned this yet)
b. Why is it not invertible?
c. Let f(x) = 3(x+2)^2 + 4, x>=-2, why is f invertible?
d. What is the domain of f^-1
e. Determine rule for f^-1 (x)

**Shyam** · October 13th 2008, 06:52 PM

$g(x)= 3(x+2)^2+4$

$y= 3(x+2)^2+4$

a) This is an upward parabola with vertex (-2,4)

See the attached graph.

b) since, $g^{-1}(x)$ is a parabola, (opening to the right) is not a function (since it does not pass the vertical line test). So, g(x) is not invertible.
domain of g(x) is R.

c) we have restricted the domain of f(x) such that its inverse $f^{-1}(x)$ become a function.

e) To find inverse of f(x), interchange x and y and solve for y.

$y= 3(x+2)^2+4$

$x= 3(y+2)^2+4$

$x-4= 3(y+2)^2$

$(y+2)^2=\frac{x-4}{3}$

$y+2=\pm \sqrt{\frac{x-4}{3}}$

$y=\pm \sqrt{\frac{x-4}{3}}-2$

$f^{-1}(x)=\pm \sqrt{\frac{x-4}{3}}-2$

d) for domain, the value inside the square root $\ge 0$

domain of $f^{-1}(x), \;\;\frac {x-4}{3}\ge 0$

$\Rightarrow x -4\ge 0$

$\Rightarrow x \ge 4$

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