If (-1, -1) is an extremum of
, then both partial derivatives vanish at this point.
Compute the gradients and evaluate them at the given point.
![\nabla f = \left\langle y - \dfrac1{x^2}, x - \dfrac1{y^2}\right\rangle \implies \nabla f (-1,-1) = \langle-2,-2\rangle \neq \langle0,0,\rangle](https://tex.z-dn.net/?f=%5Cnabla%20f%20%3D%20%5Cleft%5Clangle%20y%20-%20%5Cdfrac1%7Bx%5E2%7D%2C%20x%20-%20%5Cdfrac1%7By%5E2%7D%5Cright%5Crangle%20%5Cimplies%20%5Cnabla%20f%20%28-1%2C-1%29%20%3D%20%5Clangle-2%2C-2%5Crangle%20%5Cneq%20%5Clangle0%2C0%2C%5Crangle)
![\nabla f = \langle 2x+2,0\rangle \implies \nabla f(-1,-1) = \langle0,0\rangle](https://tex.z-dn.net/?f=%5Cnabla%20f%20%3D%20%5Clangle%202x%2B2%2C0%5Crangle%20%5Cimplies%20%5Cnabla%20f%28-1%2C-1%29%20%3D%20%5Clangle0%2C0%5Crangle)
![\nabla f = \langle y, x-2y\rangle \implies \nabla f(-1,1) = \langle-1,1\rangle \neq\langle0,0\rangle](https://tex.z-dn.net/?f=%5Cnabla%20f%20%3D%20%5Clangle%20y%2C%20x-2y%5Crangle%20%5Cimplies%20%5Cnabla%20f%28-1%2C1%29%20%3D%20%5Clangle-1%2C1%5Crangle%20%5Cneq%5Clangle0%2C0%5Crangle)
![\nabla f = \left\langle y + \frac1{x^2}, x + \frac1{y^2}\right\rangle \implies \nabla f(-1,1) = \langle0,0\rangle](https://tex.z-dn.net/?f=%5Cnabla%20f%20%3D%20%5Cleft%5Clangle%20y%20%2B%20%5Cfrac1%7Bx%5E2%7D%2C%20x%20%2B%20%5Cfrac1%7By%5E2%7D%5Cright%5Crangle%20%5Cimplies%20%5Cnabla%20f%28-1%2C1%29%20%3D%20%5Clangle0%2C0%5Crangle)
The first and third functions drop out.
The second function depends only on
. Compute the second derivative and evaluate it at the critical point
.
![f(x,y) = x^2+2x \implies f'(x) = 2x + 2 \implies f''(x) = 2 > 0](https://tex.z-dn.net/?f=f%28x%2Cy%29%20%3D%20x%5E2%2B2x%20%5Cimplies%20f%27%28x%29%20%3D%202x%20%2B%202%20%5Cimplies%20f%27%27%28x%29%20%3D%202%20%3E%200)
This indicates a minimum when
. In fact, since this function is independent of
, every point with this
coordinate is a minimum. However,
![x^2 + 2x = (x + 1)^2 - 1 \ge -1](https://tex.z-dn.net/?f=x%5E2%20%2B%202x%20%3D%20%28x%20%2B%201%29%5E2%20-%201%20%5Cge%20-1)
for all
, so (-1, 1) and all the other points
are actually <em>global</em> minima.
For the fourth function, check the sign of the Hessian determinant at (-1, 1).
![H(x,y) = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} -2/x^3 & 1 \\ 1 & -2/y^3 \end{bmatrix} \implies \det H(-1,-1) = 3 > 0](https://tex.z-dn.net/?f=H%28x%2Cy%29%20%3D%20%5Cbegin%7Bbmatrix%7D%20f_%7Bxx%7D%20%26%20f_%7Bxy%7D%20%5C%5C%20f_%7Byx%7D%20%26%20f_%7Byy%7D%20%5Cend%7Bbmatrix%7D%20%3D%20%5Cbegin%7Bbmatrix%7D%20-2%2Fx%5E3%20%26%201%20%5C%5C%201%20%26%20-2%2Fy%5E3%20%5Cend%7Bbmatrix%7D%20%5Cimplies%20%5Cdet%20H%28-1%2C-1%29%20%3D%203%20%3E%200)
The second derivative with respect to
is -2/(-1) = 2 > 0, so (-1, -1) is indeed a local minimum.
The correct choice is the fourth function.
Answer:
if she works 18 she earns 81 dollars i think
Answer:
![33\ ft](https://tex.z-dn.net/?f=33%5C%20ft)
Step-by-step explanation:
Let
y-----> the height of the telephone pole
we know that
In the right triangle of the figure
The sine of angle of 67 degrees is equal to divide the opposite side of angle of 67 degrees by the hypotenuse
![sin(67\°)=\frac{y}{36}](https://tex.z-dn.net/?f=sin%2867%5C%C2%B0%29%3D%5Cfrac%7By%7D%7B36%7D)
solve for y
![y=(36)sin(67\°)=33\ ft](https://tex.z-dn.net/?f=y%3D%2836%29sin%2867%5C%C2%B0%29%3D33%5C%20ft)
Answer:
C, 4a, 2b and 7
Step-by-step explanation: