MATH241 Lecture Notes - Lecture 12: Indeterminate Form
MATH241 - Lecture 12 - Midterm Exam 1 Review
*taken from a sample midterm*
Example
:
8) Consider the function √4 − x
A) Use the definition of the derivative to compute (x)f′
B) Find the domain of (x)f
C) Find the domain of (x)f′
D) Find the equation for the tangent line to the curve at the point on the curve(x)y=f
corresponding to x= − 5
Solution:
A) (x)f′= lim
h → 0 h
f(x + h) − f(x)
= lim
h → 0 h
−
√4− (x + h)√4−x
= lim
h → 0 (h
−
√4− (x + h)√4−x)( +
√4− (x + h)√4 − x
+
√4− (x + h) √4 −x )
= lim
h → 0
4 − (x + h) − (4 − x)
h ( + )
√4 − x − h√4 − x
= lim
h → 0
−h
h +
(√4 − x − h√4 − x)
= lim
h → 0
−1
+
√4 − x − h√4 − x
=−1
+
√4 − x√4 − x
= − 1
√
24−x
B) In order for to be defined, we need → or . The(x)f=√4 − x 04 − x≥ 4 ≥ 0 x≤ 4
domain is − , 4]( ∞
C) In order for to be defined, we need to be defined, and we need(x) −f′=1
√
24 − x√4 − x
=
√
24 − x/ 0
If , this implies that , thus the domain of is
√
24 − x= 0 x= 4 (x)f′(− , 4)∞
D) We need a slope and a point
(− )m=f′5 = −1
√
24 − (−5)
Document Summary
Math241 - lecture 12 - midterm exam 1 review. = lim h 0 f(x + h) f(x) h. 4 (x + h) 4 x h h 0 ( 4 (x + h) (4 x) 4 x h 4 x h ( 4 x 4 x ( 4 x h 4 x) h. 4 x h 4 x. In order for domain is , 4] (x) . 2 4 x to be defined, we need. 4 x to be defined, and we need. , thus the domain of f (x) is ( , 4) : we need a slope and a point m = f . 1 (x y 3 = 6. 5 y 3 = 6 x. 3) f (x) f (a) f (a) = lim h 0 f(x + h) f(x) h f(x) f(a) x a f(a + h) f(a) h lim h 0 is preferred.