Question

If $f:\mathbb{R} \to \mathbb{R}$ is a function such that $f\left( x \right) = {x^3} + {x^2}f'\left( 1 \right) + xf''\left( 2 \right) + f'''\left( 3 \right)$ for all $x \in \mathbb{R}$ , then prove that $f\left( 2 \right) = f\left( 1 \right) - f\left( 0 \right)$

Answer 1

Hint:
We can find the 1st 2nd and 3rd derivatives of the given function. Then we can find $f'\left( 1 \right)$ , $f''\left( 2 \right)$ and $f'''\left( 3 \right)$. Then we can obtain their values by solving them. Then we can rewrite the given function. Then we can find the values of $f\left( 2 \right)$ , $f\left( 1 \right)$ and $f\left( 0 \right)$. Then we can take the RHS and prove that it is equal to the LHS.

Complete step by step solution:
We have the function $f\left( x \right) = {x^3} + {x^2}f'\left( 1 \right) + xf''\left( 2 \right) + f'''\left( 3 \right)$
We can take the 1st derivative,
 $ \Rightarrow f'\left( x \right) = 3{x^2} + 2xf'\left( 1 \right) + f''\left( 2 \right) + 0$
We can find the value of $f'\left( x \right)$ at $x = 1$ . On substituting the value of x in the above equation, we get,
 $ \Rightarrow f'\left( 1 \right) = 3{\left( 1 \right)^2} + 2 \times 1 \times f'\left( 1 \right) + f''\left( 2 \right)$
On rearranging, we get,
 $ \Rightarrow f'\left( 1 \right) = 3 + 2f'\left( 1 \right) + f''\left( 2 \right)$
On adding like terms, we get,
 $ \Rightarrow - f'\left( 1 \right) = 3 + f''\left( 2 \right)$ …. (1)
Now we can take the 2nd derivative. For that we can differentiate the 1st derivative one more time.
 $ \Rightarrow f''\left( x \right) = 6x + 2f'\left( 1 \right) + 0$
We can find the value of $f''\left( x \right)$ at $x = 2$ . On substituting the value of x in the above equation, we get,
 $ \Rightarrow f''\left( 2 \right) = 6 \times 2 + 2f'\left( 1 \right)$
On simplification, we get,
 $ \Rightarrow f''\left( 2 \right) = 12 + 2f'\left( 1 \right)$ … (2)
To find the value of $f''\left( 2 \right)$ and $f'\left( 1 \right)$ we can solve the equation (1) and (2).
On substituting equation (2) in (1), we get,
 $ \Rightarrow - f'\left( 1 \right) = 3 + 12 + 2f'\left( 1 \right)$
On rearranging we get,
 $ \Rightarrow - 3f'\left( 1 \right) = 15$
On dividing throughout with -3, we get,
 $ \Rightarrow f'\left( 1 \right) = - 5$ …. (3)
Now we can substitute this in equation (2).
 $ \Rightarrow f''\left( 2 \right) = 12 + 2\left( { - 5} \right)$
On simplification we get,
 $ \Rightarrow f''\left( 2 \right) = 12 - 10$
On solving we get,
 $ \Rightarrow f''\left( 2 \right) = 2$ …. (4)
Now we can take the 3rd derivative. For that we can differentiate the 2nd derivative one more time.
 $ \Rightarrow f'''\left( x \right) = 6$
We can find the value of $f'''\left( x \right)$ at $x = 3$ . On substituting the value of x in the above equation, we get,
 $ \Rightarrow f'''\left( 3 \right) = 6$ … (5)
So, we can rewrite the function by substituting equation (3), (4) and (5) as,
 $ \Rightarrow f\left( x \right) = {x^3} + {x^2}\left( { - 5} \right) + x\left( 2 \right) + \left( 6 \right)$
On simplification we get,
 $ \Rightarrow f\left( x \right) = {x^3} - 5{x^2} + 2x + 6$
Now we can find the values of $f\left( 2 \right)$ , $f\left( 1 \right)$ and $f\left( 0 \right)$ , by substituting the values in the place of x.
When $x = 0$ , we get,
 $ \Rightarrow f\left( 0 \right) = {0^3} - 5 \times {0^2} + 2 \times 0 + 6$
On simplification we get,
 $ \Rightarrow f\left( 0 \right) = 6$ … (6)
When $x = 1$ , we get,
 $ \Rightarrow f\left( 1 \right) = {1^3} - 5 \times {1^2} + 2 \times 1 + 6$
On simplification we get,
 $ \Rightarrow f\left( 1 \right) = 1 - 5 + 2 + 6$
On adding we get,
 $ \Rightarrow f\left( 1 \right) = 4$ … (7)
When $x = 2$ , we get,
 $ \Rightarrow f\left( 2 \right) = {2^3} - 5 \times {2^2} + 2 \times 2 + 6$
On simplification we get,
 $ \Rightarrow f\left( 2 \right) = 8 - 20 + 4 + 6$
On adding we get,
 $ \Rightarrow f\left( 2 \right) = - 2$ …. (8)
 We need to prove that $f\left( 2 \right) = f\left( 1 \right) - f\left( 0 \right)$ , we can take the RHS.
 $ \Rightarrow RHS = f\left( 1 \right) - f\left( 0 \right)$
On substituting equation (6) and (7), we get,
 $ \Rightarrow RHS = 4 - 6$
 $ \Rightarrow RHS = - 2$
On substituting equation (8), we get,
 $ \Rightarrow RHS = f\left( 2 \right)$
We also have the \[LHS = f\left( 2 \right)\]
 $ \Rightarrow RHS = LHS$
Hence proved.

Note:
We must note that the value of a particular function at a given point is a constant value. Similarly, the value of the derivatives at a particular point is also a constant. Therefore, while differentiating the given function, we must consider them as constants and proceed with integration. We need to find the values of the derivative for the points given in the function so that we can simplify them and solve to find their value.
We can calculate the value of the function or derivative at a particular point by substituting the point in the place of x. $f'''\left( x \right) = 6$ will have the constant value of 6 for every value of x.