# Is #f(x)=(-3x^3+15x^2-2x+1)/(x-4)# increasing or decreasing at #x=2#?

decreasing at x = 2

Evaluate f'(a) to find out if a function f(x) is increasing or decreasing at x = a.

• At x = a, f(x) is increasing if f'(a) > 0.

• At x = a, f(x) is decreasing if f'(a) < 0.

f(x) is decreasing at x = 2 graph{(-3x^3+15x^2-2x+1)/(x-4) [-29.65, 29.67, -14.86, 14.8]} because f'(2) < 0.

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To determine if ( f(x) = \frac{-3x^3 + 15x^2 - 2x + 1}{x - 4} ) is increasing or decreasing at ( x = 2 ), we need to evaluate the derivative of ( f(x) ) at ( x = 2 ) and see if it is positive or negative.

Using the quotient rule, the derivative of ( f(x) ) is:

( f'(x) = \frac{(x - 4)(-9x^2 + 30x - 2) - (-3x^3 + 15x^2 - 2x + 1)(1)}{(x - 4)^2} )

Evaluating ( f'(2) ), we get:

( f'(2) = \frac{(2 - 4)(-9(2)^2 + 30(2) - 2) - (-3(2)^3 + 15(2)^2 - 2(2) + 1)(1)}{(2 - 4)^2} )

( f'(2) = \frac{(-2)(-36 + 60 - 2) - (-24 + 60 - 4 + 1)}{(-2)^2} )

( f'(2) = \frac{(-2)(-78) - 33}{4} )

( f'(2) = \frac{156 - 33}{4} )

( f'(2) = \frac{123}{4} )

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

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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