Hi,
I'm confused about how to find the answer to this question... I think I need a super quick refresh on how to draw a parabola and figure out turning points and stuff from an equation...
The problem in question is
Let g: [b, infinity) --> R, where g(x) = x^2 + 4x. If b is the smallest real number such that g has an inverse function find b and g^-1 (x).
I literally just put that into my CAS to find the smallest real number that b could be. (so, b is -4) but I don't remember how to do this irl without a cas....
Also, I'm not quite sure how to complete the rest of the question.
Actually, the smallest value of \(b\) is \(-2\), but let's find out why that is.
Recall that, for \(g^{-1}\), the inverse function of \(g\), to exist, \(g\)
must be a one-to-one function.
If we think about the general shape of a positive quadratic, we see that in this case, the smallest possible value of \(b\) will actually be the \(x\)-coordinate of the turning point. (See the diagram below)
The local minimum of \(g\) occurs at \(x=\dfrac{-4}{2\times 1}=-2\), so the minimal value of \(b\) is \(-2\).\[g:[-2,\infty)\to\mathbb{R},\ g(x)=x^2+4x.\] Now, let's answer the second part of the question. For convenience, let \(y=g^{-1}(x)\). \begin{align*}x&=y^2+4x\\
\implies x&=(y+2)^2-4 \tag{complete the square}\\
\implies x+4&=(y+2)^2\end{align*} Note that, when we square root both sides, we have to take the
positive root only, but why? Well, recall that \(\text{range}(g^{-1})=\text{domain}(g)=[-2,\infty)\), so \(g^{-1}(x)+2\geq0\). Thus, \[g^{-1}(x)=\sqrt{x+4}-2.\] Except, we haven't fully answered the question. Note that we are asked to find \(g^{-1}\), not just its rule, so we need to supply a domain.
From the graph, it is easy to see that \(\text{range}(g)=\text{domain}(g^{-1})=[-4,\infty)\). Thus, \[\boxed{g:[-4,\infty)\to\mathbb{R},\ g^{-1}(x)=\sqrt{x+4}-2.}\]