Quadratic in vertex form
can anyone help me find the inverse of the function f(x)= -1/2 (x-3)^2 -3 ? This is a quad in vertex form if confused.
1 Answer
We have
\[f(x)=y= -\frac{1}{2} (x-3)^2 -3 \Rightarrow -\frac{1}{2} (x-3)^2=y+3\]
\[\Rightarrow (x-3)^2=-2(y+3)\]
\[\Rightarrow x-3=\pm\sqrt{-2(y+3)}\]
\[\Rightarrow x=3\pm\sqrt{-2(y+3)}.\]
Note that thif function is not invertable in general, but if we restrict the function $f(x)$ to $x \geq 3$ or $x\leq 3$, then it will be invertable, and the inverse will be obtained by switching the roles of $x$ and $y$ in the above equation:
\[ y=3+\sqrt{-2(x+3)} \text{or} y=3-\sqrt{-2(x+3)},\]
respectively.
Paul F
200
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- 1 Answer
- 416 views
- Pro Bono
Related Questions
- Find $\lim _{x \rightarrow 0} x^{x}$
- Differentiate $f(x)=\int_{\sqrt{x}}^{\sin^2 x}\arctan (1+e^{t^2})dt$
- Factorize algebraic equation and describe the steps taken.
- Algebra problem
- Find $\lim_{x\rightarrow \infty} \frac{1}{x^2}\sin x^2\tan x$
- AP Calculus problem
- How do you prove that if $y = x^n$ then the derivative is $\frac{dy}{dx} = nx^{n-1}$
- Find the exact form (Pre-Calculus)
