First Derivative & Critical Points
For f(x)=x3−6x2+9x+4,
find all critical numbers.
x=1, 3
Find the intervals of concavity for
f(x)=x3−6x2+9x+1.
f′′(x)=6x−12
Thus, Concave down on (−∞,2) Concave up on (2,∞)
What information does the first derivative provide when sketching a curve?
It determines where the function is increasing, decreasing, and has critical points.
Find the vertical asymptote of
f(x)=(x+2)/(x-3).
x=3
A function has
f′(x)=−(x−2)2(x+1).
Without finding f(x), determine where the function is increasing and decreasing.
Increasing: (−∞,−1)
Decreasing: (−1,∞)
For f(x)=x4−4x3, determine the intervals on which the function is increasing and decreasing.
Decreasing: (−∞,3)
Increasing: (3,∞)
Find the point of inflection of
f(x)=x3−9x2+24x−5.
Answer:
f′′(x)=6x−18
Set f′′(x)=0:
x=3, f(3)=27−81+72−5=13
Therefore: (3,13)
What information does the second derivative provide when sketching a curve?
It determines concavity and helps identify possible points of inflection.
Find the horizontal asymptote of
f(x)=(3x2−1)/(x2+4).
y=3
A function satisfies f′(x)=x(x−2)(x+3).
At which critical number does the function have a local maximum?
Critical numbers: x=−3, 0, 2
The sign changes from + to − at x=0.
x=0
Find the local maximum and local minimum of
f(x)=x3−3x2−9x+5.
Local maximum (−1,10)
Local minimum (3,−22)
For f(x)=x4−4x3+10, find the intervals where the graph is concave up and concave down.
f′′(x)=12x2−24x=12x(x−2)
Concave up: (−∞,0)∪(2,∞)
Concave down: (0,2)
For f(x)=x3−3x, determine the local maximum and local minimum.
f′(x)=3x2−3=3(x−1)(x+1)
Critical numbers: x=−1, 1 f(−1)=2,f(1)=−2
Therefore: Local maximum (−1,2) Local minimum (1,−2)
Determine the vertical and horizontal asymptotes of
f(x)=(2x+5)/(x−4).
x=4, y=2
Find the equation of the tangent line to
f(x)=x3−6x2+9x+2
at its local maximum.
f′(x)=3(x−1)(x−3)
Local maximum occurs at x=1. f(1)=6
Since the tangent at a local maximum is horizontal: y=6
A function has f′(x)=x2(x−2)(x+3).
Identify its critical numbers and determine whether the function changes from increasing to decreasing or decreasing to increasing at each critical number.
Critical numbers:
x=−3, 0, 2
Sign analysis gives:
Does f′′(x)=0 always indicate a point of inflection? Explain
Answer: No. The concavity must actually change at the point.
For f(x)=x4−2x2, find all critical points and classify them.
f′(x)=4x3−4x=4x(x−1)(x+1)
Critical numbers: x=−1,0,1
Values: f(−1)=−1,f(0)=0,f(1)=−1
Therefore: Local minima (−1,−1),(1,−1) Local maximum (0,0)
For f(x)=(x2−1)/(x−1),
identify any discontinuity and determine whether it is a vertical asymptote.
There is a hole at (1,2)
There is no vertical asymptote
Determine the number of local extrema of
f(x)=x5−5x3+4x.
f′(x)=5x4−15x2+4
Let u=x2: 5u2−15u+4=0
u=(15±1451/2)/10
Both values are positive, producing four distinct critical numbers:
The derivative changes sign at each, so there are
4 local extrema
Given f′(x)=(x−1)2(x+2), determine all local extrema of f(x).
Critical numbers:
x=−2, x=1
At x=−2, f′ changes from negative to positive:
Local minimum at x=−2
At x=1, f′ does not change sign:
No local extremum at x=1
Determine all points of inflection of
f(x)=x5−5x4+5x3.
Answer:
f′′(x)=20x3−60x2+30x =10x(2x2−6x+3)
Thus: x=0, x=(31/2−3)/2, x=(31/2+3)/2
All three produce changes in concavity, so these are the inflection x-values.
Sketch the general shape of f(x)=x4−4x2.
Identify its critical points, local extrema, and points of inflection.
Critical numbers: x=−21/2, 0, 21/2
Values: f(±21/2)=−4, f(0)=0
So: Local minima (±21/2,−4) Local maximum (0,0)
Also, f′′(x)=12x2−8
f′′(x)=0⇒x=±32
The points of inflection are
(−(2/3)1/2, −20/9),((2/3)1/2,−20/9)
Determine the vertical and horizontal asymptotes of
f(x)=(2x3−5x2+x+3)/(x2−1).
Vertical asymptote: x=−1
There is a hole at x=1.
Polynomial division gives: f(x)=(2x−5)+[2/(x+1)]
Therefore, the function has a slant asymptote:
y=2x−5
Perform a complete curve analysis of
f(x)=x2/(x-1)
Domain: x2=1
Intercept: x2=0⇒(0,0)
Vertical asymptote: x=1
Slant asymptote: y=x+1
Critical numbers: x=0, 2
Increasing: (−∞,0)∪(2,∞)
Decreasing: (0,1)∪(1,2)
Function values: f(0)=0,f(2)=4
Therefore: Local maximum (0,0) Local minimum (2,4)
Concave down: (−∞,1)
Concave up: (1,∞)
Concavity changes at x=1, but f(1) is undefined, so there is no actual point of inflection.