HOW TO FIND THE POINTS WHERE TANGENT LINE IS HORIZONTAL

For each of the following curves, find the point at where the tangent line drawn to the curve is horizontal.

Example 1 :

y = x² - 4x - 12 

Solution :

y = x² - 4x - 12 ----(1)

Slope of the tangent line :

ᵈʸ⁄dₓ = 2x - 4

Since the tangent line is horizontal, slope is equal to zero.

ᵈʸ⁄dₓ = 0

2x - 4 = 0

2x = 4

x = 2

Substitute x = 2 into (1).

y = 2² - 4(2) - 12

y = 4 - 8 - 12

y = -16

(x, y) = (2, -16)

Example 2 :

y = x³ - 18x + 5 

Solution :

y = x³ - 18x + 5 ----(1)

Slope of the tangent line :

ᵈʸ⁄dₓ = 3x² - 18

Since the tangent line is horizontal, slope is equal to zero.

ᵈʸ⁄dₓ = 0

3x² - 18 = 0

3x² = 18

x² = 9

x = ±√9

x = ±3

x = -3  or  x = 3

Substitute x = -3 and x = 3 into (1).

When x = -3,

y = x³ - 18x + 5

y = (-3)³ - 18(-3) + 5

y = -27 + 54 + 5

y = 32

When x = 3,

y = x³ - 18x + 5

y = 3³ - 18(3) + 5

y = 27 - 54 + 5

y = -22

(x, y) = (-3, 32) and (3, -22)

Example 3 :

y = x⁴ - 2x² + 3

Solution :

y = x⁴ - 2x² + 3 ----(1)

Slope of the tangent line :

ᵈʸ⁄dₓ = 4x³ - 4x

Since the tangent line is horizontal, slope is equal to zero.

ᵈʸ⁄dₓ = 0

4x³ - 4x = 0

4x(x² - 1) = 0

4x(x² - 1²) = 0

4x(x + 1)(x - 1) = 0

4x = 0  or  x + 1 = 0  or  x - 1 = 0

x = 0  or  x = -1  or  x = 1

Substitute x = 0, x = -1 and x = 1 into (1).

When x = 0,

y = 0 - 0 + 3

y = 3

When x = -1,

y = (-1)⁴ - 2(-1)²+ 3

y = 1 - 2(1) + 3

y = 1 - 2 + 3

y = 2

When x = 1,

y = (1)⁴ - 2(1)²+ 3

y = 1 - 2(1) + 3

y = 1 - 2 + 3

y = 2

(x, y) = (0, 3), (-1, 2) and (1, 2)

Example 4 :

y = x + sinx,  0 ≤ x ≤ 2π

Solution :

y = x + sinx ----(1)

Slope of the tangent line :

ᵈʸ⁄dₓ = 1 + cosx

Since the tangent line is horizontal, slope is equal to zero.

ᵈʸ⁄dₓ = 0

1 + cosx = 0

cosx = -1

x = π ∈ 0 ≤ x ≤ 2π

Substitute x = π into (1).

y = π + sinπ

y = π + 0

y = π

(x, y) = (π, π)

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