What is the average speed of a commercial
2 answers:
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Compute the gradient of .
Set this equal to the zero vector and solve for the critical points.
The last case has no real solution, so we can ignore it.
Now,
so we have two critical points (0, 0) and (2, 2).
Compute the Hessian matrix (i.e. Jacobian of the gradient).
Check the sign of the determinant of the Hessian at each of the critical points.
which indicates a saddle point at (0, 0);
We also have , which together indicate a local minimum at (2, 2).
Answer:
The value of the expression increases as j decreases
Step-by-step explanation:
Let
As j decreases, the value of j300 decreases (i.e the farther j300 is from 150). Due to the wider gap between 150 and j300, the value of f(j) increases.
For example:
When j = 1, f(j) = 150 - (300*1) = -150
When j = 0.5, f(j) = 150 - (300*0.5) = 0
When j = 0. f(j) = 150 - (300*0) = 150
It is obvious from the analogy above that the expression 150-j150−j150 increases as j decreases