Question

2. If f(x)=2x^2-5x, find the values of k if f^-1(3)=k
@iambatman @wio

Answer 1

Start off by finding the inverse of f(x)

Answer 2

Remember that: \[
f^{-1}(f(x)) = x
\]

Answer 3

So when \(x=k\), we can say \(f(x)=3\).

Answer 4

Which also means that \(f(k)=3\)

Answer 5

Yeah, that's good

Answer 6

I tried by i got \[f^-1(x)=\sqrt{2x+10}\]

Answer 7

but*

Answer 8

\[f(k)=2k^2-5k=3\]

Answer 9

So you need to solve for \(k\) here. It's a quadratic equation.

Answer 10

\[
2k^2-5k-3=0
\]

Answer 11

k=5 and k=3/2

Answer 12

Do you know how to solve quadratic equations

Answer 13

Your inverse function is wrong,
\[f(x)=2x^2-5x\]

The inverse of this function is \[f^{-1} (x)= \frac{ 1 }{ 4 } (5 \pm \sqrt{8x+25})\]
\[f^{-1} (3) = \frac{ 1 }{ 4 }(5\pm \sqrt{8(3)+25})\]
\[f^{-1} (3) = \frac{ 1 }{ 4 }(5 \pm 7) = k\]

Answer 14

Here is how you would find and inverse: \[
f(x) =ax^2+bx+c\\
0=ax^2+bx+c-f(x)
\]Then you would use quadratic equation, with \(c-f(x)\) substituted in for what would be \(c\).\[
x = \frac{-b\pm\sqrt{b^2-4a(c-f(x))}}{2a}
\]By definition \(f^{-1}(f(x))=x\), and we clearly have an expression of \(f(x)\) that equals \(x\), the only problem is that the \(\pm\) means it is not a function. We have two options:\[
f^{-1}(x)=\frac{-b\color{red}+\sqrt{b^2-4a(c-x)}}{2a}\\
f^{-1}(x)=\frac{-b\color{red}-\sqrt{b^2-4a(c-x)}}{2a}
\]

Answer 15

thnx a lot @iambatman @wio