# First find f' and then find f. f''(x) = 2x^3+3x^2+10 f'(1)=-9 f(1)=-5 f(x)=? f(x)=?

## Question:

First find {eq}f' {/eq} and then find {eq}f. {/eq}

{eq}f''(x) = 2x^3+3x^2+10 \\ f'(1)=-9 \\ f(1)=-5 \\ f(x)=? \\ f(x)=? {/eq}

## Double Derivative:

If f(x) is a function which is differentiated once, it will bring out f'(x). it is also known as the first derivative of the function. Now the first derivative of the function is differentiated again to produce {eq}f''(x) {/eq} which is also known as the second derivative. Also, integral of a function is also termed as the antiderivative. It can be expressed mathematically as follows:

{eq}\int f''(x) \ dx = f'(x) + C \\ \int f'(x) \ dx = f(x) + C {/eq}

where 'C' is the constant of integration.

Given:

{eq}f''(x) = 2x^3 + 3x^2 + 10 \\ f'(1) = -9 \\ f(1) = -5 \\ {/eq}

Integrating the f(x):

{eq}\int f''(x) = \int (2x^3 + 3x^2 + 10) \ dx \\ \Rightarrow f'(x) = 2 ( \dfrac {x^4}{4} ) + 3 ( \dfrac {x^3}{3}) + 10 x + C_1 \\ \Rightarrow f'(x) = ( \dfrac {x^4}{2} ) + x^3 + 10 x + C_1 \\ {/eq}

On applying the condition of f'(x):

{eq}\Rightarrow f'(1) = \dfrac {1^4}{2} + 1^3 + 10(1) + C_1 \\ \Rightarrow -9 = \dfrac {1}{2} + 1 + 10 + C_1 \\ \Rightarrow -9 = \dfrac {1}{2} + 11 + C_1 \\ \Rightarrow C_1 = \dfrac {-41}{2} \\ {/eq}

On putting the value of {eq}C_1 {/eq} in f'(x) equation:

{eq}f'(x) = \dfrac {x^4}{2} + x^3 + 10 x - \dfrac {41}{2} \\ {/eq}

Integrating equation of f'(x):

{eq}\int f'(x) = \int (\dfrac {x^4}{2} + x^3 + 10 x - \dfrac {41}{2} ) \ dx \\ \Rightarrow f(x) = \dfrac {1}{2} (\dfrac {x^5}{5}) + \dfrac {x^4}{4} + 10 x^2 - \dfrac {41}{2} x + C_2 \\ \Rightarrow f(x) = \dfrac {x^5}{10} + \dfrac {x^4}{4} + 10 x^2 - \dfrac {41}{2} x + C_2 \\ {/eq}

On applying the condition of f(x) i.e. f(1) = -5:

{eq}\Rightarrow f(1) = \dfrac {1^5}{10} + \dfrac {1^4}{4} + 10 (1^2) - \dfrac {41}{2} (1) + C_2 \\ \Rightarrow -5 = \dfrac {1}{10} + \dfrac {1}{4} + 10 - \dfrac {41}{2} + C_2 \\ \Rightarrow -5 -10 = \dfrac {1}{10} + \dfrac {1}{4} - \dfrac {41}{2} + C_2 \\ \Rightarrow -15 = -20.15 + C_2 \\ \Rightarrow C_2 = 5.15 \\ {/eq}

On putting the value of {eq}C_2 {/eq} in the equation of f(x):

{eq}\Rightarrow f(x) = \dfrac {x^5}{10} + \dfrac {x^4}{4} + 10 x^2 - \dfrac {41}{2} x + 5.15 \\ {/eq}