# HOW TO FIND THE POINTS WHERE TANGENT LINE IS HORIZONTAL

The slope of an horizontal line is always zero. Let us consider the curve given by the function y = f(x). To find the slope of a tangent line to y = f(x), we have to find the first derivative of the function y = f(x), that is ᵈʸ⁄d.

ᵈʸ⁄d represents the slope of a tangent line to the curve y = f(x).

If the tangent line is horizontal, then its slope is equal to zero.

ᵈʸ⁄d = 0

To find the point at where the tangent line is horizontal, equate the slope ᵈʸ⁄d to zero and solve for x. Substitute the value of x into y = f(x) and find the value of y. Write the point (x, y) at where the tangent line to the curve 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⁄d = 2x - 4

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

ᵈʸ⁄d = 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⁄d = 3x2 - 18

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

ᵈʸ⁄d = 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⁄d = 4x3 - 4x

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

ᵈʸ⁄d = 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⁄d = 1 + cosx

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

ᵈʸ⁄d = 0

1 + cosx = 0

cosx = -1

x = π ∈ ≤ x ≤ 2π

Substitute x = π into (1).

y = π + sinπ

y = π + 0

y = π

(x, y) = (π, π)

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