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1. Find the derivative of <span>P(x)=3x^3+2x^2-6x. It's P'(x)=9x^2 + 4x - 6.
2. Set this result equal to zero and solve for the critical values:
</span> 9x^2 + 4x - 6 = 0 Using the quadratic formula, I got
x = [-4 plus or minus sqrt(232)] / 18. Reducing this,
x = [-4 plus or minus 2 sqrt(58)]; thus, there are two real, unequal roots and two real, unequal critical values.
3. One at a time, examine the two critical values: determine whether the derivative changes from neg to pos or from pos to neg at each of these values. Example: If the derivative is pos to the left of the first c. v. and neg to the right, we've got a local max.
4. Since there are only 2 critical values, you can have no more than 1 local max (corresponding to a change in the sign of the derivative from pos to neg) and one local min. (from neg to pos).
Message me if this explanation is not sufficient to help you understand this problem thoroughly.
So even just multiplying 3x2 by 6x we know it has to be D because it’s the inly one with 18x3 but i went ahead and worked out the whole problem using the box method
it’s in the attached picture
Answer: =
Step-by-step explanation:
Answer:
<h3>√(1-a)²+(b+2)²</h3>
Step-by-step explanation:
If the pipe on the plan runs from point N(a, -2) to point P(1, b), the expression that represents the shortest distance between N and P in units, is expressed using the formula;
D = √(x₂-x₁)²+(y₂-y₁)²
Given the coordinates N(a, -2) and P(1, b);
x₁ = a, y₁ = -2, x₂ = 1 and y₂ = b
NP = √(1-a)²+(b+2)²
<em>Hence the expression that represents the shortest distance between N and P in units is √(1-a)²+(b+2)² units</em>
