Find the best value of the derivative function
The author did not @guodongLovesOi derivative usage specification, if the fight is not responsible teacher of mathematics
For the first quadratic function \ (f (x) = x ^ 2 + 3x + 1 \)
We can easily obtain the derivative:
\ [set \ triangle = \ lim _ {\ triangle-> 0}, x '= x- \ triangle \\ then f' (x) = \ frac {f (x) -f (x ')} {x- x'} \\ i.e. \\ \ frac {x ^ 2 + 3x + 1- {x '} ^ 2-3x'-1} {x-x'} = \\ \ frac {(x + x ') ( x-x') + 3 (x-x ')} {x-x'} = \\ i.e. \ quad {2x + 3} \
] set point coordinates \ (( V, F (V)) \) , can be seen from the figure, \ (f '(v) = 0 \)
\(\therefore v=-\frac{3}{2}\)
The \ (v \) substitution:
\ [(- \ frac {3 } {2}, {(- \ frac {3} {2})} ^ 2-3 \ times \ frac {3} {2} +1) \\ (- \ frac { 3} {2}, - \
frac {5} {4}) \] by checking cute apex above formula may be right.
Then extended to a quadratic function \ (f (X) = AX + BX + C ^ 2 \)
\ [set \ triangle = \ lim _ {\ triangle-> 0}, x '= x- \ triangle \\ so f ( x) the derivative f '(x) can be calculated as: \\ \ frac {ax ^ 2 + bx + ca {x'} ^ 2-bx'-c} {x-x '} = \\ \ frac {a (x ^ 2- (x ') ^ 2) + b (x-x')} {x-x '} = \\ \ frac {a (x + x') (x-x ') + b (x -x ')} {x-x '} = \\ a (x + x ') + b = \\ i.e. \ quad 2ax + b \]
into the above quadratic function can be pretty simple checking this is correct .
\ [2AV + B = V = 0 \\ - \ FRAC {B} {}. 2A \\ V (- \ FRAC {B} {}. 2A, F (- \ FRAC {B} {}. 2A)) \]
Grass, like textbooks conclusion, the X can not be installed.
Informatics section
Next we extend to \ (n \) order function. (Template: Rule of Thirds )
The process is easy to find Derivative
\(f'(x)=\frac{f(x)-f(x')}{x-x'}\)
However, a cancer of \ (n \) linear function, the more difficult as we transformed as above
But we can be certain a specific point of it!
Consider the title of \ (l, r \) constraints, we just need to find the closest \ (0 \) in that position just fine.
A water problem