# The Derivatives Of Piecewise Functions

## Mathematics lesson, subject derivatives. Finding of derivatives of the piecewise functions. Analysing the continuity and derivative of the functions at critical points. The subject expression and solved questions.

**The Piecewise Functions**

The functions defined in different ways in different number ranges are called piecewise functions.

If we wants to find the derivative of piecewise functions, we must be analysis whether or not continuity in this point.

If the function is continious in the critical points, we examine the left and right derivatives of the function. If the right and left derivatives are equal, the function have a derivative at the criticial points.

If the derivative of the function is asked at a point other than the critical points, the derivative is taken in the region where the function is defined.

For a function to have derivative at one point, the function must be continuous at that point.

**Example:**

f(x) = 5x^{3} + 1, if x > 3

f(x) = 2x^{2} – 1, if ≤ 3

Describe whether the function f(x) has a derivative at point x = 3 .

**Solution:**

The function f(x) is defined in two way for the point x = 3, the cases that the x>3, and the cases that x smaller or equal to 3.

If we come to the value 3 from right side,

We use the function f(x) = 5x^{3} + 1

If we come to the value 3 from left side.

We use the function f(x) = 2x^{2} -1 .

Let we examine the continuity of the function f(x) at the point x= 3.

Lim_{x→3+} (5x^{3} + 1)

= 5.27 + 1 = 136

Lim_{x→3-} (2x^{2} – 1)

= 18 – 1 = 17

Lim_{x→3+} f(x) ≠ Lim_{x→3-}

The function f(x) not continiuty at the point x = 3, so it has not a derivative.

**Example:**

f(x) = x^{2} + 6, if x ≥ 4

f(x) = 6x – 2, if x< 4

1- Examine the derivative of the function f(x) at point x = 4.

2- Examine the derivative of the function f(x) at point x = -4.

3- Examine the derivative of the function f(x) at point x = 6.

**Solution:**

1-

Let we analysis the left and right limits for examine the continuity of the function f(x).

Limx→4+ (x^{2} + 6) = 16 + 6 = 22

= 22

Lim_{x→4-} (6x – 2 ) = 24 – 2

= 22

The left and right limits are equal. Therefore the function f(x) is continuous at the point x = 4.

Now, let we examine the right and left derivatives of the function.

if x ≥ 4

f(x) = x^{2} + 6

f’(x) = 2x + 6

f’(4) = 14

if x < 4 ,

f(x) = 6x – 2

f’(x) = 6

f’(4) = 6

Since the left and right derivatives are not equal the function has no a derivative at this point.

2-

The function f(x) has been defined in the form f(x) = 6x – 2 for x = - 2

Lim_{x→-1+} (6x – 2) = - 8

Lim_{x→-1-} (6x – 2) = - 8

The left and right limits are equal, therefore the function is continuous at this point.

f’(x) = 6

f’( - 2) = 6

3-

The function f(x) has been defined in the form f(x) = x^{2} + 6 for x = 6

Lim_{x→6+} (x^{2} + 6) = 42

Lim_{x→6-} (x^{2} + 6) = 42

The left and right limits are equal.

f’(x) = 2x

f’(6) = 12

**Example:**

f(x) = 5x^{3} – 3, if x ≥ 1

f(x) = 3x^{5} – 1, if x < 1

1- Examine the derivative of the function f(x) at point x = 1.

2 – Examine the derivative of the function f(x) at point x = 2.

3- Examine the derivative of the function f(x) at point x = -2.

**Solution:**

1-

Let we analysis the continuity of the function f(x) at point x = 1.

Lim_{x→1+} (5x^{3} – 3) = 2

Lim_{x→1-} (3x^{5} – 1) = 2

The function f(x) is continuous at the point x = 1.

For x ≥ 1 ,

f(x) = 5x^{3} – 3

f’(x) = 15x^{2}

f’(1) = 15

For x < 1,

f(x) = 3x^{5} – 1

f’(x) = 15x

f’(1) = 15

The left and right derivatives of the function are equal, therefore the function f(x) has a derivative at the point x = 1.

2-

The function f(x) has been defined in the form f(x) = 5x^{3} – at the point x = 2

Lim_{x→2+} (5x^{3} – 3) = 37

Lim_{x→2-} (5x^{3}– 3) = 37

The left and right limits are equal.

f’(x) = 15x^{2}

f’(2) = 60 dir.

3-

The function f(x) has been defined in the form 3x^5 – 1 at the point x = -2

Lim_{x→-2+} (3x^{5} – 1) = - 95

Lim_{x→-2-} (3x^{5} – 1) = - 95

The left and right limits are equal, therefore the function f(x) is continuous at this point.

f’(x) = 15x^{4}

f’( - 2) = 240

**Derivative Of Parametric Functions**

**Derivative Of The Sum Of Functions**

RISE KNOWLEDGE

25/03/2019

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