Q.If $f : R \to R$, $f(x) = x^2 + 5x + 9$, then find the value of $f^{-1}(8)$ and $f^{-1}(9)$.

Solve $f(x)=8$ and $f(x)=9$ as quadratic equations in $x$.

Given $f(x)=x^2+5x+9$.

Step 1: $f^{-1}(8)$. Put $f(x)=8$:

$x^2+5x+9=8 \;\Rightarrow\; x^2+5x+1=0$

By the quadratic formula, $x=\dfrac{-5\pm\sqrt{25-4}}{2}=\dfrac{-5\pm\sqrt{21}}{2}$.

Step 2: $f^{-1}(9)$. Put $f(x)=9$: …

Solve $f(x)=1$ directly: $x^2-5x+7=1$ factors to $(x-2)(x-3)=0$, giving $x=2$ or $x=3$.

Given $f(x) = x^2-5x+7$. We need the value(s) of $x$ such that $f(x)=1$, i.e. $f^{-1}(1)$.

$x^2-5x+7=1$

$x^2-5x+6=0$

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

So $x=2$ or $x=3$.

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