A point that satisfies f′(x)=0f'(x)=0 is called a Stationary Point
A sign diagram is a way of visualizing a curve’s graph by indicating the inflection points, stationary points, and the increases or decreases in the curve.
First, equate f′(x)f'(x) to 00 to find the stationary points
f(x)f(x)
==
x4-8x2+16x4−8x2+16
f′(x)f'(x)
==
4x3-16x=04x3−16x=0
Equate f′(x)f'(x) to 00
==
4x(x2-4)=04x(x2−4)=0
Factor out 4x4x
==
4x(x-2)(x+2)=04x(x−2)(x+2)=0
Factor out the difference of 22 squares
x=0,x=2,x=-2x=0,x=2,x=−2
Find the corresponding yy values by substituting each xx value to the function
f(x)f(x)
==
x4-8x2+16x4−8x2+16
f(0)f(0)
==
04−8(02)+1604−8(02)+16
==
0-0+160−0+16
==
1616
Stationary Point: (0,16)(0,16)
f(x)f(x)
==
x4-8x2+16x4−8x2+16
f(2)f(2)
==
24−8(22)+1624−8(22)+16
==
16-32+1616−32+16
==
00
Stationary Point: (2,0)(2,0)
f(x)f(x)
==
x4-8x2+16x4−8x2+16
f(−2)f(−2)
==
−24−8(−22)+16−24−8(−22)+16
==
16-32+1616−32+16
==
00
Stationary Point: (−2,0)(−2,0)
Next, create a sign diagram to identify where the curve increases or decreases
Start by setting up a horizontal line with matching indicators of the stationary points
Test the gradient of a point to the left of -2−2 such as x=-3x=−3
f′(x)f'(x)
==
4x3-16x4x3−16x
f′(−3)f′(−3)
==
4(−3)3−16(−3)4(−3)3−16(−3)
==
4(-27)+484(−27)+48
==
-108+48−108+48
==
-60−60
This value is negative, which means the curve’s slope at -3−3 is decreasing
Indicate this on the sign diagram by adding a negative sign to the left of -2−2
Test the gradient of a point to the right of -2−2 such as x=-1x=−1
f′(x)f'(x)
==
4x3-16x4x3−16x
f′(−1)f′(−1)
==
4(−1)3−16(−1)4(−1)3−16(−1)
==
4(-1)+164(−1)+16
==
-4+16−4+16
==
1212
This value is positive, which means the curve’s slope at -1−1 is increasing
Indicate this on the sign diagram by adding a positive sign to the right of -2−2
Test the gradient of a point to the left of 22 such as x=1x=1
f′(x)f'(x)
==
4x3-16x4x3−16x
f′(1)f′(1)
==
4(13)−16(1)4(13)−16(1)
==
4(1)-164(1)−16
==
4-164−16
==
-12−12
This value is negative, which means the curve’s slope at 11 is decreasing
Indicate this on the sign diagram by adding a negative sign to the left of 22
Test the gradient of a point to the right of 22 such as x=3x=3
f′(x)f'(x)
==
4x3-16x4x3−16x
f′(3)f′(3)
==
4(33)−16(3)4(33)−16(3)
==
4(27)-484(27)−48
==
108-48108−48
==
6060
This value is positive, which means the curve’s slope at 33 is increasing
Indicate this on the sign diagram by adding a positive sign to the left of 22
Finally, draw a curve along the stationary points with the help of the sign diagram
Question 4 of 5
4. Question
To help identify where the curve increases or decreases, sketch the first derivative of the curve: