Critical points in a curve include the Global Minimum and Maximum (also called Absolute Minimum and Maximum) and Local Minimum and Maximum (also called Relative Minimum and Maximum)
(a) Find the Global Minimum from the given curve.
A global minimum has the lowest function value for all x in the domain
Hence, the Global Minimum is Point A.
(b) Find the Global Maximum from the given curve.
A global maximum has the highest function value for all x in the domain
[highlight only point F in green]
Hence, the Global Minimum is Point F.
(c) Find the Local Minimum from the given curve.
A local minimum is a stationary or turning point that has the lowest function value for all x in a small interval around it. This is not applicable for endpoints or boundary points
[highlight only point D in green]
Hence, the Local Minimum is Point D.
(d) Find the Local Maximum from the given curve.
A local maximum is a stationary or turning point that has the highest function value for all x in a small interval around it. This is not applicable for endpoints or boundary points
[highlight only point B in green]
Hence, the Local Minimum is Point B.
(a) Global Minimum: A
(b) Global Maximum: F
(c) Local Minimum: D
(d) Local Maximum: B
Question 2 of 2
2. Question
Find the critical points of the curve f(x)=3x2+12x−5 under the domain −4≤x≤4
Write coordinates in the format “(a,b)”
Write 0 if the answer does not exist
Critical points in a curve include the Global Minimum and Maximum (also called Absolute Minimum and Maximum) and Local Minimum and Maximum (also called Relative Minimum and Maximum)
A point that satisfies f'(x)=0 is called a Stationary Point
First, find the stationary point of the curve by equating its first derivative to 0
f(x)
=
3x2+12x−5
f'(x)
=
6x+12=0
Equate f'(x) to 0
6(x+2)=0
Factor out 6
x=−2
This means that a stationary point exists when x=−2
Find the corresponding y value of the stationary point by solving for f(−2)
f(x)
=
3x2+12x−5
f(−2)
=
3(−22)+12(−2)−5
Substitute x=−2
=
3(4)−24−5
=
12−24−5
y
=
−17
Hence, a stationary point is at (−2,−17)
Next, confirm if this stationary point is a minimum or maximum by solving for f”(x)
f(x)
=
3x2+12x−5
f'(x)
=
6x+12
f”(x)
=
6
This value is positive, which means the stationary point (−2,−17) is a minimum point
Now, find the two boundary points by solving for f(−4) and f(4) (from −4≤x≤4)
f(x)
=
3x2+12x−5
f(−4)
=
3(−4)2+12(−4)−5
=
3(16)−48−5
=
48−48−5
=
−5
A boundary point is at (−4,−5)
f(x)
=
3x2+12x−5
f(4)
=
3(4)2+12(4)−5
=
3(16)+48−5
=
48+48−5
=
91
A boundary point is at (4,91)
Form the curve by connecting the points, then identify critical points
The Global Minimum has the lowest function value for all x in the domain
Hence, it is the point (−2,−17)
The Global Maximum has the highest function value for all x in the domain
Hence, it is the point (4,91)
The Local Minimum is a stationary or turning point that has the lowest function value for all x in a small interval around it. This is not applicable for endpoints or boundary points
Hence, it is the point (−2,−17)
The Local Maximum is a stationary or turning point that has the highest function value for all x in a small interval around it. This is not applicable for endpoints or boundary points
Enter the corresponding letter for the point
