4.1
The equation for linearization.
What is L(x)=f’(a)(x-a)+f(a)
The reason why extrema may not exist on an interval.
What is a discontinuity or open interval.
Conditions for application of MVT.
What is the function must be continuous and differentiable on interval [a, b].
The point where concavity changes.
What is an inflection point.
what is represented by f(x), f'(x), and f''(x)
f(x)=s(t)=position
f'(x)=v(t)=velocity
f''(x)=a(t)=acceleration
The cube root of 27 is 3. How much larger is the cube root of 27.2? Estimate using linear approximation.
f(27+.2) - f(27)= 0.00739
f'(y)dy = 1/3(x)-2/3 dy = 1/3(27)-2/3 (.2) = 0.00741
I0.00739-0.00741I= 0.00002
Determine whether Rolle's theorem applies to the following function on the given interval then, if so, find the points guaranteed to exist:
f(x)=sin(2x) [0, pi/2]
continuous: yes
differentiable: yes
f(0)=0 f(pi/2)=0
f'(x)=2cos(2x)=0
x= pi/4 and 3pi/4
Use the MVT to determine the values of c for:
f(x)=(x-2)2 on [1, 4]
f'(x)=2(x-2)
f(1)=(1-2)2 f(4)=(4-2)2=4
(4-1)/(4-1)=1
2(c-2)=1 c=(5/2) f'(5/2)=1
Find the first and second derivative of the following:
f(x)=ln(2x+4)
f'(x)=1/(x+2)
f''(x)=-1/(x+2)2
v(t)=6x3+2x2-468x
a(t)=18x2+4x-468
How accurate is L(a+.1) ≈ f(a+.1)?
f(x)=ln(x+1) a=0
f(0)=ln(1)=0 f'(0)=1/1=1
L(0.1)/0.1=.1
f(0.1)/ln(1/1)=.0953
I.1-0.0953I=.0046
Find the critical points for the function:
f(x)=(4-x2 )1/2
f'(x)=-x/(4-x2 )1/2
-x=0 x=0
(4-x2 )1/2 =0 x=+/- 2
Use the first derivative test to find intervals where increasing and decreasing of:
f(x)=x2/(24-3x)
f'(x)=(48x-9x2)/(24-3x)2
f'(x)=0 x=0, 16/3 cp: 0, 16/3, 8
Test values: -1(-), 1(+), 6(-), 9(-)
Decreasing: (-inf, 0), (16/3, 8), (8, inf)
Increasing: (0, 16/3)
Use the second derivative test to find the relative extrema of the following:
f(x)= x1/2-2cosx
f'(x)= 21/2+2sinx=0 x= 3pi/4 and 7pi/4
f''(x)=2cosx f''(3pi/4)=- f''(7pi/4)=+
Relative max: 7pi/4
Relative min: 3pi/4
Find the total distance traveled by the particle for the following function from [0, 2]:
f(x)=x3-6x2+9x
f'(x)=3x2-12x+9 x=1 and 3
f''(x)=6x-12 x=2 crit points on [0,2]=0, 1, 2
f(0)=0 f(1)=4 f(2)=2
------0-----2-----4------
total distance: 4
Calculate the linearization of: f(x)= 1/(2+x)1/2 at a=2
L(x)=f'(a)(x-a)+f(a)
f'(2)=-1/2(2+x)3/2 =-1/16
(-1/16)(x+2)+(1/2)=(-1/16)x + 3/8
Find the extreme values of the function on the given interval:
f(x)=x2/3-2x1/3 on [-1, 3]
f'(x)=2x-1/3-2/3x-2/3 cp: 0, 1
f(-1)=3 abs. max
f(0)=0
f(1)=-1 abs. min
f(3)=-0.804 local min
Find the intervals where increasing and decreasing and locate all relative extrema for:
f(x)= (1/2)x-sinx
f'(x)= (1/2)-cosx=0
x= pi/3 and 5pi/3
Test values: pi/4(-), pi/2(+), 7pi/4(-)
Increasing: (pi/3, pi/5)
Decreasing: (-inf, pi/3) and (5pi/3, inf)
Find the intervals of concavity and inflection points of the following function:
f(x)=t3((6-x)2)1/2
f'(x)=(18-3x)/3(6-x)1/3 f''(x)=(10x-72)/9(6-x)4/3
f''(x)=0 x=7.2 and 6
f''(5)=- f''(7)=- f''(8)=+
conc down: (-inf, 6) and (6, 7.2) conc up: (7.2, inf)
Inflection point: (7.2, 8.131)
Find the v(t), a(t), displacement, and total distance traveled for the following function on the given interval:
s(t)=t3-9t2+24t+20 on [1.5, 7]
v(t)=3t2-18t+24 a(t)=6t-18
s(7)-s(1.5)=90-39.125=50.875
v(t)=0 t=2 and 4
s(1.5)=39.125 s(2)=40 s(4)=36 s(7)=90
---36---39.125---40---90---
0.875+4+54=58.875
Calculate the linearization of f(x)=sin(ln(2x)) at a=pi
f'(x)=cos(ln(2x))/x
f(pi)=0.965 f'(pi)=-0.084
-0.084(x-pi)+0.965=-0.084x-1.229
Find the extrema of the function:
f(x)=(2x2)/(x-1)
f'(x)=(2x2-4x)/(x-1)2
2x2-4x=0 x=0,2 (x-1)2=0 x=1
------0------1-------2------
f'(-1)=1.5 f'(.5)=-6 f'(1.5)=-6 f'(3)=15
Increasing, decreasing, decreasing, increasing
Local max: x=0 Local min: x=2
Find the intervals where increasing and decreasing of the following function:
f(x)=(x2-3)/(x-2)
f'(x)=(x2-4x+3)/(x-2)2
(x-3)(x-1)/(x-2)2 x=3, 1, 2
asymptote at x=2
Increasing: (-inf, 1) and (3, inf)
Decreasing: (1, 2) and (2, 3)
Identify the domain, y', y'', cp, intervals where increasing decreasing, max/min values, p.o.i, and concavity for the following:
f(x)=5x2/5-2x
Domain: (-inf, 0) u (0, inf)
f'(x)=2x-3/5-2 x=1
f''(x)=-6/5x-8/5
Increasing: - Decreasing: (-inf, 1) u (1, inf)
No min or max, no inflection points
Given that s(t)=x-cos(x2)/2, find v(t), a(t) and the total distance traveled from [0, 2].
v(t)=tsin(t2) a(t)=2t2cos(t2)+sin(t2)
v(t)=0 t=0, pi1/2, -(pi1/2)
s(0)=0 s(pi1/2)=1.386 s(2)=1.327
---0---1.327---1.386---
1.386+0.059=1.445