Chain Rule
Implicit Derivatives
Mean Value Theorem
?
1st d/dx Test 4 Local max/mins
100
ex
ex
100
dy/dx (y)
equals dy/dx
100
What are the conditions of the Mean Value Theorem?
Continuous over [a,b], differentiable on (a,b)
100
If there is an open interval containing c on which f(c) is a maximum, then f(c) is called what? What is f(c) when the open interval is a minimum?
When f(c) is a maximum f(c) is a relative maximum, and when f(c) is a minimum f(c) is a relative minimum
100
if f' changes from positive to negative as x passes through c, then f has a local ?
Max at c
200
y=cos(2x)
answer: -2sin(2x)
200
y = 3√(2x − 5)
dy/dx = (2/3)* (2x − 5)-(2/3)
200
I slope of the f'(c) line, the secant line or the tangent line?
Tangent line
200
What are the condtions of the Extreme Value Theorem?
f(x) must be continuous over [a,b]
200
if f' changes from negative to positive as x passes through c, then f' has a local ?
Min at c
300
ln(2x3 − 8x5)
(3-20x2)/(x-4x3)
300
x + xy − 2x3 = 2
dy/dx = −2x-2 + 4x
300
What is the Mean Value Theorem equation?
f'(c) = (f(b)-f(a)) / (b-a)
300
f(−2)=−5 and f'(-2)=-9 ′
if x equals -2 what is the equation of the tangent line?
y - 2 = 9(x-5) or y + 5 = 9(x+2)
y + 5 = 9(x+2)
300
if f' does not change sign at c (so that f' is positive on both sides of c or negative on both sides of c), then f ?
Neither local Max/MIn
400
cos2(x2)
2(cos(x2))(-sin(x2)(2x) or -4x(sin(x2))(cos(x2))
400
x3y−2x3+y4=8
dy/dx = (4x−3x2y) / (x3+4y3)
400
g(x) = √(5x-1) and let c be the number that satisfies the Mean Value Theorem for g on the interval [1,10]. What is the value of c?
C = 4.25
g′(c)= (f(10)−f(1)) / (10−1)
f(10)−f(1)) / (10−1)=((7−2) / 9) = (5/9)
g'(x) = (5/9) = 4.25
400
The tangent line of the function f(x) at the point (1,5) passes through the point (3,4)
what is f'(x)
slope = (4-5) / (3-1)
slope = -1/2
f'(x) = -(1/2)
400
Part 1)
f(x) = (x2-1)2
Identify all critical points
x = 0,1,-1
500
((3x3+2x2)/(x))2
36x3+36x2+8x
500
y = 3√(2 + tan (x2))
dy/dx = (2/3) x sec2 (x2 )(2 + tan (x2))-(2/3)
500
f(x)= √(4x-3) and let c be the number that satisfies the Mean Value Theorem for f on the interval 1 ≤ x ≤ 3. what is c?
C = 1.75
f'(c) = (f(3)−f(1)) / ((3)-(1)) = ((3-1) / 2) = 1
f'(x) = (2 / (√(4x-3))
x = 1.75
500
y=7
y'=1
x=-4
y−7=1(x+4)
y=x+11
500
Part 2)
f(x) = (x2-1)2
Find the intervals where the function is increasing or decreasing
Increasing: [-1,0] u [1, ∞)
Decreasing: (∞,-1] u [0,1]