How do you find the critical points for #g(x)=2x^3-3x^2-12x+5#?
Find the first derivative and set the equation to equal zero.
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To find the critical points of ( g(x) = 2x^3 - 3x^2 - 12x + 5 ), follow these steps:
- Calculate the derivative of ( g(x) ) with respect to ( x ), denoted as ( g'(x) ).
- Set ( g'(x) ) equal to zero and solve for ( x ).
- The solutions obtained in step 2 are the critical points of ( g(x) ).
First derivative of ( g(x) ): [ g'(x) = 6x^2 - 6x - 12 ]
Setting ( g'(x) ) equal to zero and solving for ( x ): [ 6x^2 - 6x - 12 = 0 ] [ 2x^2 - 2x - 4 = 0 ]
Now, solve this quadratic equation. You can use methods like factoring, completing the square, or the quadratic formula.
Once you find the values of ( x ), those are the critical points of ( g(x) ).
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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.
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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