Question-108864

Question Number 108864 by I want to learn more last updated on 19/Aug/20

Commented by I want to learn more last updated on 19/Aug/20

What is the maximum length of PQ

Answered by mr W last updated on 19/Aug/20

f (x) = x^{3} - 3 x - 2

g (x) = 2 x - 2

f^{'} (x) = 3 x^{2} - 3

s u c h t h a t P Q i s m a x i m u m, t a n g e n t

a t Q m u s t b e p a r a l l e l t o g (x) .

f^{'} (x) = 3 x^{2} - 3 = g^{'} (x) = 2

\Rightarrow x = \pm \sqrt{\frac{5}{3}}

a t x = - \sqrt{\frac{5}{3}} :

y_{P} = - 2 \sqrt{\frac{5}{3}} - 2

y_{Q} = - \frac{5}{3} \sqrt{\frac{5}{3}} + 3 \sqrt{\frac{5}{3}} - 2

P Q = - \frac{5}{3} \sqrt{\frac{5}{3}} + 3 \sqrt{\frac{5}{3}} - 2 - (- 2 \sqrt{\frac{5}{3}} - 2)

= \frac{10 \sqrt{15}}{9}

a t x = \sqrt{\frac{5}{3}} :

y_{P} = 2 \sqrt{\frac{5}{3}} - 2

y_{Q} = \frac{5}{3} \sqrt{\frac{5}{3}} - 3 \sqrt{\frac{5}{3}} - 2

P Q = 2 \sqrt{\frac{5}{3}} - 2 - (\frac{5}{3} \sqrt{\frac{5}{3}} - 3 \sqrt{\frac{5}{3}} - 2)

= \frac{10 \sqrt{15}}{9}

\Rightarrow {P Q}_{m a x, l o c a l} = \frac{10 \sqrt{15}}{9}

{P Q}_{m a x} = \infty w h e n x \to \pm \infty

Commented by I want to learn more last updated on 25/Aug/20

Thanks sir, I appreciate

Answered by Aziztisffola last updated on 19/Aug/20

P (x; 2 x - 2) and Q (x; x^{3} - 3 x - 2)

PQ (x) = \sqrt{0^{2} + {(x^{3} - 3 x - 2 - 2 x + 2)}^{2}}

= x^{3} - 5 x

PQ {(x)}^{'} = {3 x}^{2} - 5

PQ {(x)}^{'} = 0 \Rightarrow {3 x}^{2} - 5 = 0 \Rightarrow x = \underline{+} \sqrt{\frac{5}{3}}

\min in x = \sqrt{\frac{5}{3}}

Max local in x = - \sqrt{\frac{5}{3}}; look at variation of PQ

then PQ (- \sqrt{\frac{5}{3}}) = {(- \sqrt{\frac{5}{3}})}^{3} + 5 \sqrt{\frac{5}{3}}

= \frac{10 \sqrt{15}}{9}

Commented by I want to learn more last updated on 25/Aug/20

Thanks sir, i appreciate