Question:
Let \(f:[0,\infty)\to\mathbb{R}\) be defined by
\[
f(x)=\frac{3x^2+4x+1}{x^2+3x+2}.
\]
Then the value of \((f^{-1})'(2)\) is:
Let \(f:[0,\infty)\to\mathbb{R}\) be defined by
\[
f(x)=\frac{3x^2+4x+1}{x^2+3x+2}.
\]
Then the value of \((f^{-1})'(2)\) is:
Show Hint
To find \((f^{-1})'(a)\), first solve \(f(x)=a\). Then compute \(f'(x)\) at that point and take the reciprocal.
Updated On: Jun 11, 2026
- \(15\)
- \(5\)
- \(0\)
- \(25\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
If \(y=f(x)\), then
\[
(f^{-1})'(y)=\frac1{f'(x)}
\]
where \(x=f^{-1}(y)\).
Step 1: Find \(x\) such that \(f(x)=2\).
\[ \frac{3x^2+4x+1}{x^2+3x+2}=2. \] Cross multiplying, \[ 3x^2+4x+1=2x^2+6x+4. \] \[ x^2-2x-3=0. \] \[ (x-3)(x+1)=0. \] Since domain is \([0,\infty)\), \[ x=3. \] Thus \[ f^{-1}(2)=3. \]
Step 2: Differentiate \(f(x)\).
Using quotient rule, \[ f'(x)= \frac{(6x+4)(x^2+3x+2)-(3x^2+4x+1)(2x+3)} {(x^2+3x+2)^2}. \] Substitute \(x=3\). Numerator: \[ (22)(20)-(40)(9) =440-360 =80. \] Denominator: \[ 20^2=400. \] Hence \[ f'(3)=\frac{80}{400} =\frac15. \]
Step 3: Apply inverse derivative formula.
\[ (f^{-1})'(2) = \frac1{f'(3)} = \frac1{\frac15} = 5. \] Hence \[ \boxed{5}. \]
Step 1: Find \(x\) such that \(f(x)=2\).
\[ \frac{3x^2+4x+1}{x^2+3x+2}=2. \] Cross multiplying, \[ 3x^2+4x+1=2x^2+6x+4. \] \[ x^2-2x-3=0. \] \[ (x-3)(x+1)=0. \] Since domain is \([0,\infty)\), \[ x=3. \] Thus \[ f^{-1}(2)=3. \]
Step 2: Differentiate \(f(x)\).
Using quotient rule, \[ f'(x)= \frac{(6x+4)(x^2+3x+2)-(3x^2+4x+1)(2x+3)} {(x^2+3x+2)^2}. \] Substitute \(x=3\). Numerator: \[ (22)(20)-(40)(9) =440-360 =80. \] Denominator: \[ 20^2=400. \] Hence \[ f'(3)=\frac{80}{400} =\frac15. \]
Step 3: Apply inverse derivative formula.
\[ (f^{-1})'(2) = \frac1{f'(3)} = \frac1{\frac15} = 5. \] Hence \[ \boxed{5}. \]
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