Week 4 Course Notes - MATH 137 WEEK 4 NOTES PROF DOUG PARK...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 137 WEEK 4 NOTES PROF. DOUG PARK 1. Definition of derivative L x ( x , f ( x )) ( t , f ( t )) x t h y = f ( x ) C Let t = x + h . Line L has slope = f ( t )- f ( x ) t- x = f ( x + h )- f ( x ) h . As t → x , or equivalently as h → 0, line L approaches tangent line C at ( x,f ( x )). [slope of L ]-→ [slope of tangent line C ] Def = f ( x ) Def . The derivative of f at x is the slope of tangent line to graph y = f ( x ) at the point ( x,f ( x )), and equals f ( x ) = lim t → x f ( t )- f ( x ) t- x = lim h → f ( x + h )- f ( x ) h [Memorize!] If either limit exists as a finite number, then we say f is differentiable at x . Remark . All h ’s and t ’s will disappear after successfully evaluating the above limits. Notation . f ( x ) = ( f ( x )) = d dx f ( x ) = df dx = dy dx = y = ˙ y = ··· Physics . s ( t ) = position of particle at time t s ( t ) = velocity of particle at time t s 00 ( t ) = ( s ( t )) = acceleration of particle at time t Date : October 9, 2009. 1 2 DOUG PARK 2. Computing derivatives using limit definition Ex . Let f ( x ) = 3 √ x = x 1 / 3 . (i) If x 6 = 0, find f ( x ) using the definition. (ii) Show that f (0) DNE. (iii) Show that y = 3 √ x has a vertical tangent line at x = 0. Sol’n . (i) Recall A 3- B 3 = ( A- B )( A 2 + AB + B 2 ). f ( x ) = lim t → x f ( t )- f ( x ) t- x = lim t → x 3 √ t- 3 √ x t- x = lim t → x 3 √ t- 3 √ x ( 3 √ t ) 3- ( 3 √ x ) 3 = lim t → x 1 ( 3 √ t ) 2 + 3 √ t · 3 √ x + ( 3 √ x ) 2 If we replace t by x , then the denominator is not 0, because x 6 = 0 by assumption. Hence f ( x ) = 1 3( 3 √ x ) 2 = 1 3 · 1 x 2 / 3 = 1 3 x- 2 / 3 . (ii) Note t 2 / 3 = ( t 1 / 3 ) 2 = square ≥ 0. f (0) = lim t → f ( t )- f (0) t- = lim t → t 1 / 3 t = lim t → 1 t 2 / 3 = 1 + = ∞ So f (0) DNE (as a real number). (iii) Theorem . Let f be a continuous function. Then y = f ( x ) has a vertical tangent line at x = a ⇐⇒ lim x → a | f ( x ) | = ∞ . We check lim x → | f ( x ) | = lim x → 1 3 · 1 x 2 / 3 = lim x → 1 3 · 1 x 2 / 3 = 1 3 · 1 + = ∞ Done! Ex . Decide whether f , g are differentiable at x = 0. f ( x ) = ( x sin π x if x 6 = 0 if x = 0 g ( x ) = ( x 2 sin π x if x 6 = 0 if x = 0 Sol’n . f (0) = lim t → f ( t )- f (0) t- = lim t → t sin ( π t )- t = lim t → sin π t Note that lim t → ± π t = ±∞ , hence by letting u = π t f (0) = lim u →±∞ sin u = DNE, so f is not differentiable at x = 0. g (0) = lim t → g ( t )- g (0) t- = lim t → t 2 sin ( π t )- t = lim t → t sin π t MATH 137 3 Since- 1 ≤ sin π t ≤ 1 by multiplying by t ,- t ≤ t sin π t ≤ t if t >- t ≥ t sin π t ≥ t if t < In both cases, lim t → ( ± t ) = 0, so by Squeeze Theorem, lim t → t sin π t = 0 = g (0) Thus...
View Full Document

{[ snackBarMessage ]}

Page1 / 11

Week 4 Course Notes - MATH 137 WEEK 4 NOTES PROF DOUG PARK...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online