Q.Let $f$ be an invertible real function. Find the value of $(f^{-1}\circ f)(1)+(f^{-1}\circ f)(2)+(f^{-1}\circ f)(3)+(f^{-1}\circ f)(4)$.

To find the inverse, solve $y = f(x)$ for $x$ in terms of $y$, then relabel.

Given $f(x) = 6x - 5$. Let $y = f(x) = 6x-5$.

Solve for $x$: $6x = y+5 \Rightarrow x = \dfrac{y+5}{6}$.

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$f^{-1}(0)$ is the pre-image of $0$: solve $2x-3=0$ to get $x=\dfrac32$.

Finding a pre-image using the inverse of a function is a CBSE/NCERT Class 12 relations and functions idea.

By definition $f^{-1}(0)$ is the value of $x$ for which $f(x)=0$:

$2x - 3 = 0 \Rightarrow 2x = 3 \Rightarrow x = \frac{3}{2}.$

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f(x) = 3^x is a bijection from R to (0, ∞); its inverse is the logarithm base 3.

Let y = f(x) = 3^x. To find f^{-1}, solve for x in terms of y:

$3^x = y \Rightarrow x = \log_3 y$

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f(x)=x^2 is not one-one on Q, so f^{-1}(9) is the SET of all x with x^2=9, which is {3,-3}.

We need every $x\in Q$ such that $f(x)=x^2=9$. Solving, $x=\pm 3$. Since $f$ is a many-one function (both $x$ and $-x$ give the same square), $f^{-1}(9)$ is not a single val …

To invert $f$, set $y=f(x)$, solve for $x$ in terms of $y$, then relabel.

Let $y = f(x) = 5x+4$. Solve for $x$: $y-4 = 5x \Rightarrow x = \dfrac{y-4}{5}$.

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solve y=f(x) for x in terms of y

Let $y=f(x)=(3-x^3)^{1/3}$.

$y^3=3-x^3\implies x^3=3-y^3\implies x=(3-y^3)^{1/3}$

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