Let f - R - R be f x - x 3- x 2- x -1 and g x - f -1 x - then g -2 isA- 2B- 6C- 1-2D- 1-6

The correct option is D 1/6We have, g(x) = f^ 1(x) ⇒ f(g(x)) = x Differentiating both sides w.r.t. x, we get ⇒ f'(g(x)) × g'(x) = 1 f'(x) = 3x^2 + 2x + 1 Let ...

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Standard XII

Mathematics

Chain Rule of Differentiation

Let f : R → R...

Question

Let $f : R \to R be f (x) = x^{3} + x^{2} + x - 1$ and $g (x) = f^{- 1} (x)$ , then $g^{'} (2)$ is

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\frac{1}{6}

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Solution

The correct option is C $\frac{1}{6}$
We have, $g (x) = f^{- 1} (x) \Rightarrow f (g (x)) = x$
Differentiating both sides w.r.t. x, we get
$\Rightarrow f^{'} (g (x)) \times g^{'} (x) = 1$
$f^{'} (x) = 3 x^{2} + 2 x + 1$

$Let g (2) = K$
$We know f (g (2)) = 2$
$\Rightarrow f (K) = 2$
$\Rightarrow K^{3} + K^{2} + K - 1 = 2$
$\Rightarrow K^{3} + K^{2} + K - 3 = 0$
$∴ K = 1 \Rightarrow g (2) = 1.$
$g^{'} (x) = \frac{1}{f^{'} (g (x))}$
$g^{'} (2) = \frac{1}{f^{'} (g (2))} = \frac{1}{f^{'} (1)} = \frac{1}{6}$
$∵ f^{'} (1) = 6$

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Let f:R→R be f(x)=x3+x2+x−1 and g(x)=f−1(x), then g′(2) is

Let $f : R \to R be f (x) = x^{3} + x^{2} + x - 1$ and $g (x) = f^{- 1} (x)$ , then $g^{'} (2)$ is