Find f(x) given that f(3) = -2 and f'(3)=4.
Using the reminder theorem yields:
f(x) = (x - 3)*q(x) + (-2)
Differentiating the function with respect to x yields:
f'(x) = q(x) + (x-3)*q'(x)
Plugging x = 3 in the equation of derivative yields:
f'(3) = q(3)
Hence, the derivative of the function at x = 3 equals the value of the quotient q(x) at x = 3.
Supposing that the order of the polynomial function is n then the order of the polynomial quotient must be n-1. (the divisor is a binomial of first order).
f(x) = a_n*x^n + ......+a_0
f'(x) = q(x) => f'(3) = q(3) => q(3) = 4
Hence, if n = 1 => f(x) = mx + b => f(3) = 3m + b = -2
f'(x) = m => f'(3) = 4 => m = 4 => 12 + b = -2 => b = -14
Hence, if n = 2 => f(x) = ax^2 + bx + c
f(3) = 9a + 3b + c = -2
f'(x) = 2ax + b => f'(3) = 6a + b = 4
Under these circumstances, three variables and just two equations, the second order function may not be determined.
Hence, under the circumstances f(3)=-2 and f'(3)=4, the function may be fairly determined only if it is a linear function with the equation y -4x + 14 = 0.
It is given that f'(3) = 4 and f(3) = -2, using which the function f(x) has to be determined.
There could be many functions of different orders of x that satisfy these conditions.
Assuming the function f(x) has an order 1, f(x) = ax + b
f(3) = a*3 + b = -2
f'(x) = a, the value for f'(x) is a constant for all values of x.
f'(3) = a = 4
Substituting a = 4 in 3a + b = -2
=> 12 + b = -2
=> b = -14
The function f(x) = 4x - 14