HOW TO FIND THE FUNCTION FROM THE DERIVATIVE

Differentiation and integration are opposite process. To find the particular function  from the derivation, we have to integrate the function.

Question 1 :

If f'(x)  =  4x - 5 and f(2)  =  1, find f(x)

Solution :

f'(x)  =  4x - 5 

∫f'(x)  =  ∫(4x - 5) dx

=  (4x²/2) - 5x + c 

f(x)  =  2x² - 5x + c ----(1)

Given that :

f(2)  =  1

Instead of x, let us apply 2 in (1)

f(2)  =  2(2)² - 5(2) + c 

1  =  2 (4) - 10 + c

1  =  8 - 10 + c

1  =  -2 + c

c  =  1 + 2

c = 3

By applying the value of c in (1)

f(x)  =  2x² - 5x + 3

Hence the required function is 2x² - 5x + 3.

Question 2 :

If f'(x)  =  9x² - 6x and f(0)  =  -3, find f(x)

Solution :

f'(x)  =  9x² - 6x

∫f'(x)  =  ∫(9x² - 6x) dx

=  (9x³/3) - (6x²/2) + c 

f(x)  =  3x³ - 3x² + c ----(1)

Given that :

f(0)  =  -3

Instead of x, let us apply 0 in (1)

f(0)  =  3(0)³ - 3(0)² + c

-3  =  0 - 1 + c

-3  =  c

c = -3

By applying the value of c in (1)

f(x)  =  3x³ - 3x² - 3

Hence the required function is 3x³ - 3x² - 3.

Question 3 :

If f''(x)  =  12x - 6 and f(1)  =  30, f'(1)  =  5 find f(x)

Solution :

f''(x)  =  12x - 6

∫f''(x)  =  ∫(12x - 6) dx

=  (12x²/2) - 6x + c 

f'(x)  =  6x² - 6x + c₁ ----(1)

∫f'(x)  =  ∫(6x² - 6x + c₁) dx

∫f'(x)  =  ∫(6x² - 6x + 5) dx

f'(1)  =  5

Apply x = 1 in (1)

f'(1)  =  6(1)² - 6(1) + c₁

5  =  6 - 6  + c₁

c₁  =  5

f(x)  =  (6x³/3) - (6x²/2) + 5x + c₂

f(x)  =  2x³ - 3x² + 5x + c₂   -------(2)

f(1)  =  30

Apply x = 1 in (2)

f(1)  =  2(1)³ - 3(1)² + 5(1) + c₂ 

30  =  2 - 3 + 5 + c₂ 

30  =  4 + c₂ 

c₂ =  30 - 4

c₂ =  26

By applying c₂  =  26 in (2), we get

f(x)  =  2x³ - 3x² + 5x + 26

Hence the required function is f(x) = 2x³ - 3x² + 5x + 26.

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