What is the maximum number of relative extrema contained in the graph of this function f(x)=3x^3-x^2+4x-2

2 answers:

5 0

The maximum number of turning points in a cubic function is 2.

In this case,

$f(x)=3x^3-x^2+4x-2\implies f'(x)=9x^2-2x+4$

The discriminant is $(-2)^2-4(9)(4)=-140$ , which means the derivative has no real roots. This means there are no critical points and thus no turning points/relative extrema.

sergeinik [125]3 years ago

5 0

Answer:

the answer is 2 (apex)

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Evaluate -1-3-(-9)+(-5)

babunello [35]

To evaluate this expression, we need to remember that subtracting a negative number is the same as adding a positive number, and that adding a negative number is the same as subtracting a positive number. Using this knowledge, let's begin to simplify the expression below:

-1 - 3 - (-9) + (-5)

Because addition of a negative number is the same as subtraction of a positive number, we can change + (-5) to -5, as shown below:

-1 - 3 - (-9) - 5

Next, because we know that subtracting a negative number is the same as adding a positive number, we can change - (-9) to + 9, as shown below:

-1 - 3 + 9 - 5

Now, we can subtract the first two terms and begin to evaluate our expression:

-4 + 9 - 5

Next, we can add the first two numbers of the expression:

5 - 5

Now, we can subtract our last two numbers, which gives us our answer:

Therefore, your answer is 0.

Hope this helps!