Which of the following has the same value as
2 answers:
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Check for critical points within the unit ball by solving for when the first-order partial derivatives vanish:
Taken together, we find that (0, 0, 0) appears to be the only critical point on
within the ball. At this point, we have 
.
Now let's use the method of Lagrange multipliers to look for critical points on the boundary. We have the Lagrangian
with partial derivatives (set to 0)
We then observe that
So, ignoring the critical point we've already found at (0, 0, 0),
So ultimately, we have 9 critical points - 1 at the origin (0, 0, 0), and 8 at the various combinations of
, at which points we get a value of either of 
, with the maximum being the positive value and the minimum being the negative one.
Taken together, we find that (0, 0, 0) appears to be the only critical point on
Now let's use the method of Lagrange multipliers to look for critical points on the boundary. We have the Lagrangian
with partial derivatives (set to 0)
We then observe that
So, ignoring the critical point we've already found at (0, 0, 0),
So ultimately, we have 9 critical points - 1 at the origin (0, 0, 0), and 8 at the various combinations of
9514 1404 393
Answer:
- y-intercept: (0, -6)
- x-intercepts: (-3, 0), (-1, 0), (1, 0)
Step-by-step explanation:
We notice the first pair of coefficients is the same as the last pair (with the sign changed). This means we can factor by grouping.
f(x) = (2x^3 +6x^2) -(2x +6)
f(x) = 2x^2(x +3) -2(x +3)
f(x) = 2(x^2 -1)(x +3) = 2(x -1)(x +1)(x +3)
The factors are made to be zero when x is 1, -1, or -3.
The x-intercepts are (1, 0), (-1, 0), (-3, 0).
The y-intercept is the constant, -6.
