Is #f(x)=(2x^3-7x^2+3x+1)/(x-3)# increasing or decreasing at #x=2#?

Question

Emma Aldridge · Accepted Answer

The function is increasing at #x=2#. See explanation

To find out if a function #f(x)# is increasing or decreasing at a given point #x_0# you can calculate the first derivative #f'(x_0)#. If the derivative #f'(x_0)# is greater than zero, then the function is increasing, if it is negative, then the function is decreasing. Here the derivative can be calculated using the quotient rule: #f(x)=(2x^3-7x^2+3x+1)/(x-3)# #f'(x)=((2x^3-7x^2+3x+1)'(x-3)-(2x^3-7x^2+3x+1)(x-3)')/(x-3)^2# #f'(x)=((6x^2-14x+3)(x-3)-(2x^3-7x^2+3x+1))/(x-3)^2# #f'(x)=(6x^3-32x^2+45x-9-2x^3+7x^2-3x-1)/(x-3)^2# #f'(x)=(4x^3-25x^2+42x-10)/(x-3)^2# Now we calculate the value of #f'(2)# #f'(2)=(4*2^3-25*2^2+42*2-10)/(2-3)^2# #f'(2)=32-100+84-10=6# #f'(2)# is greater than zero, so #f(x)# is increasing at #x=2#

Axel Alvarez · Answer

undefined To determine if f(x) = (2x^3 - 7x^2 + 3x + 1) / (x - 3) is increasing or decreasing at x = 2, we need to examine the sign of the derivative of f(x) at x = 2. After computing the derivative, we evaluate its sign at x = 2. If the derivative is positive, f(x) is increasing at x = 2; if negative, f(x) is decreasing at x = 2.

The derivative of f(x) is given by f'(x) = (2x^3 - 13x^2 + 42x - 40) / (x - 3)^2.

Substituting x = 2 into the derivative, we get f'(2) = (2(2)^3 - 13(2)^2 + 42(2) - 40) / (2 - 3)^2 = (16 - 52 + 84 - 40) / 1 = 8.

Since f'(2) = 8 > 0, f(x) is increasing at x = 2.