How do you find the critical points of #b'(x)=x^3+3x^2-4x-12#?

Answer 1

Christopher Allers

#x=-3" and " x=+-2#

#"find the critical points by equating " b'(x)" to zero"#

#rArrx^3+3x^2-4x-12=0#

#rArrx^2(x+3)-4(x+3)=0#

#rArr(x+3)(x^2-4)=0#

#x+3=0rArrx=-3larr" is a critical point"#

#(x-2)(x+2)=0rArrx=+-2larr" are critical points"#

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Answer 2

Emma Aldridge

To find the critical points of the function $b'(x) = x^3 + 3x^2 - 4x - 12$ , you need to find the values of $x$ where the derivative equals zero or is undefined.

First, find the derivative of $b(x)$ by differentiating term by term: $b'(x) = \frac{d}{dx}(x^3 + 3x^2 - 4x - 12)$
Set $b'(x)$ equal to zero and solve for $x$ : $x^3 + 3x^2 - 4x - 12 = 0$
Once you find the values of $x$ that satisfy this equation, those are the critical points of the function.

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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.