# Find f(x) given that f(3) = -2 and f'(3)=4.

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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**