If-g-x-x-x-and-lim-x-2-f-x-f-2-x-2-ax-b-4-3-find-the-value-of-f-g-1-

Question Number 106309 by bemath last updated on 04/Aug/20

If g (x) = x + \sqrt{x} and \lim_{x \to 2} \frac{f (x) - f (2)}{x^{2} + ax + b} = \frac{4}{3}

find the value of {(f \circ g)}^{'} (1) .

Answered by bobhans last updated on 05/Aug/20

g (x) = x + \sqrt{x} \Rightarrow g^{'} (x) = 1 + \frac{1}{2 \sqrt{x}}; g^{'} (1) = \frac{3}{2}

g (1) = 2

{(f \circ g)}^{'} (1) = g^{'} (1) f^{'} (g (1)) = \frac{3}{2} f^{'} (2)

(1) 2^{2} + 2 a + b = 0 \to b = - 4 - 2 a

(2) \lim_{x \to 2} \frac{f (x) - f (2)}{x^{2} + ax - 2 a - 4} = \frac{4}{3}

\lim_{x \to 2} \frac{f^{'} (x)}{2 x + a} = \frac{4}{3}

case (1) if a \neq - 4, then we get

f^{'} (2) = \frac{16 + 4 a}{3} . so {(f \circ g)}^{'} (1) = g^{'} (1) . f^{'} (g (1))

= \frac{3}{2} . f^{'} (2) = \frac{3}{2} . \frac{16 + 4 a}{3} = 8 + 2 a

case (2) if a = - 4, then \lim_{x \to 2} \frac{f^{'} (x)}{2 x + a} is form \frac{0}{0}

so f^{'} (2) = 0 \Rightarrow {(f \circ g)}^{'} (1) = \frac{3}{2} . 0 = 0

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