The function f is such that f(x)=a^2x^2-ax+3b for x<1/(2a) Where a and b are constant for the case where a=1 and b=-1 Find f^-1(c.f. and find its domain I know domain of f^-1(x)=range of f(x) and it is -13/4 but I don't know inequality sign direction?

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Answer 1 · 2018-05-15T21:36:35

See below.

$a^2x^2-ax+3b$

$x^2-x-3$

Range:

Put into form $y=a(x-h)^2+k$

$h=-b/(2a)$

$k=f(h)$

$h=1/2$

$f(h)=f(1/2)=(1/2)^2-(1/2)-3=-13/4$

Minimum value $-13/4$

This occurs at $x=1/2$

So range is $(-13/4,oo)$

$f^(-1)(x)$

$x=y^2-y-3$

$y^2-y-(3-x)=0$

$y=(-(-1)+-sqrt((-1)^2-4(1)(-3-x)))/2$

$y=(1+-sqrt(4x+13))/2$

$f^(-1)(x)=(1+sqrt(4x+13))/2$

$f^(-1)(x)=(1-sqrt(4x+13))/2$

With a little thought we can see that for the domain we have the required inverse is:

$f^(-1)(x)=(1-sqrt(4x+13))/2$

With domain:

$(-13/4,oo)$

Notice that we had the restriction on the domain of $f(x)$

$x<1/2$

This is the x coordinate of the vertex and the range is to the left of this.