Is #f(x)=(-x^3+x^2-3x-4)/(4x-2)# increasing or decreasing at #x=0#?

Answer 1

#"f(x) is increasing at x=0"#

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

# • " If f'(a) > 0 , then f(x) is increasing at x = a"#
#• " If f'(a) < 0 , then f(x) is decreasing at x = a"#
To differentiate f(x) use the #color(blue)"quotient rule"#
#" If " f(x)=(g(x))/(h(x))" then"#
#color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=(h(x).g'(x)-g(x).h'(x))/(h(x))^2)color(white)(2/2)|)))#
#g(x)=-x^3+x^2-3x-4rArrg'(x)=-3x^2+2x-3#
#"and " h(x)=4x-2rArrh'(x)=4#
#rArrf'(x)#
#=((4x-2)(-3x^2+2x-3)-(-x^3+x^2-3x-4).4)/(4x-2)^2#
#rArrf'(0)=((-2)(-3)-(-4)(4))/(-2)^2#
#=22/4=11/2#

The graph{(-x^3+x^2-3x-4)/(4x-2) [-20, 20, -10, 10]} shows that f(x) is increasing at x = 0 since f'(0) > 0.

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Answer 2

To determine if ( f(x) = \frac{{-x^3 + x^2 - 3x - 4}}{{4x - 2}} ) is increasing or decreasing at ( x = 0 ), we can evaluate the sign of the derivative of ( f(x) ) at that point.

Taking the derivative of ( f(x) ) with respect to ( x ) yields:

[ f'(x) = \frac{{12x^2 - 4x - 2}}{{(4x - 2)^2}} ]

Now, substituting ( x = 0 ) into ( f'(x) ) gives:

[ f'(0) = \frac{{-2}}{{(2)^2}} = -\frac{1}{2} ]

Since ( f'(0) = -\frac{1}{2} ), which is negative, ( f(x) ) is decreasing at ( x = 0 ).

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Answer from HIX Tutor

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.

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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