What is the equation of the tangent line of f(x)=1x2−3x+2 at x=0?

p(x)=x2−3x+2⇒f=p−12f′(x)=−12p−32(2x−3)f′(0)=−122−32(−3)=+3⋅2−52=aThe desired line is g(x)=ax+bg(0)=f(0)=12=b

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Caerswso1pc

2022-03-22

What is the equation of the tangent line of

f (x) = \sqrt{\frac{1}{x^{2} - 3 x + 2}}

at x=0?

LiaisyAciskriuu

Beginner2022-03-23Added 10 answers

p (x) = x^{2} - 3 x + 2 \Rightarrow f = p^{- \frac{1}{2}}

f^{'} (x) = - \frac{1}{2} p^{- \frac{3}{2}} (2 x - 3)

f^{'} (0) = - \frac{1}{2} 2^{- \frac{3}{2}} (- 3) = + 3 \cdot 2^{- \frac{5}{2}} = a

The desired line is g(x)=ax+b

g (0) = f (0) = \sqrt{\frac{1}{2}} = b

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