Find the expression for function which has a 2nd derivative of f"(x) = 6x + 6 which passes through point (1, -2) with a slope value of 2 at that pointStep by step process

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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The function to be found has a second derivative f''(x) = 6x + 6

Integrating f''(x),

f'(x) = Int[ f''(x) dx] = Int[6x + 6 dx]

=> 3x^2 + 6x + C1

f(x) = Int[ f'(x) dx] = Int [ 3x^2 + 6x + C dx]

=> x^3 + 3x^2 + C1*x + C2

As the function has a slope of 2 at (1, -2)

3*1^2 + 6*1 + C1*x = -2

=> 3 + 6 + C1*x = -2

=> 9 + C1*1 = -2

=> C1 = -11

As the function passes through (1 , -2)

1^3 + 3*1^2 - 11*1 + C2 = -2

=> 1 + 3 - 11 + C2 = -2

=> 4 - 7 + C2 = -2

=> C2 = 5

The required function is f(x) = x^3 + 3x^2 - 11x + 5

higgsboson's profile pic

higgsboson | High School Teacher | (Level 1) eNoter

Posted on

 slope dy/dx=2

f''(X)=6x+6

first we convert it into 1st derivative equation by taking integral

so we get

dy/dx=6x^2/2+ 6x + C

dy/dx=3x^2+6x + C......eq 1 

here c is a constant .

put values  of dy/dx and x in eq 1, so we get

2=3(1)^2+6(1) + C

2=3+6 + C

therefore C=-7

now eq 1 becomes

dy/dx=3x^2+6x-7

by taking again integral we get

y=3x^3/3 + 6x^2/2 -7x +c

y=x^3+3x^2 -7x + c

now put y= -2 , x= 1

-2= (1)^3+3(1)^2 -7(1) +c

-2=1+3-7+c

c=1

the final eq is

y=x^3+3x^2-7x +1

 

 

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