Given:
![p= \frac{RT}{vm} + \frac{a+bT}{vm^{2}} \\or\\v= \frac{1}{p}[ \frac{R}{m} + \frac{a+bT}{m^{2}}]](https://tex.z-dn.net/?f=p%3D%20%5Cfrac%7BRT%7D%7Bvm%7D%20%2B%20%5Cfrac%7Ba%2BbT%7D%7Bvm%5E%7B2%7D%7D%20%5C%5Cor%5C%5Cv%3D%20%5Cfrac%7B1%7D%7Bp%7D%5B%20%5Cfrac%7BR%7D%7Bm%7D%20%20%2B%20%5Cfrac%7Ba%2BbT%7D%7Bm%5E%7B2%7D%7D%5D%20)
Answer:
When p is held constant,
![\frac{\partial v}{\partial T} |_{p}= \frac{1}{p}[ \frac{R}{m}+ \frac{b}{m^{2}}]](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20T%7D%20%7C_%7Bp%7D%3D%20%5Cfrac%7B1%7D%7Bp%7D%5B%20%5Cfrac%7BR%7D%7Bm%7D%2B%20%5Cfrac%7Bb%7D%7Bm%5E%7B2%7D%7D%5D%20%20%20)
Answer:
When p is held constant,
The figure below shows a brick in the shape of a rectangular prism. 6 cm 8 cm 14 cm (a) Use the calculator to find the surface a
1 answer:
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Answer:
Critical points are 1 and -1
Maximum at x=1
Minimum at x=-1
Step-by-step explanation:
We are given that a function
on (
)
We have to find the critical points of the function.
To find the critical point we will differentiate function w.r.t x and then substitute f'(x)=0
Therefore, the critical points of the given function are 1 and -1.
When we goes from 0 to 2 then the sign of derivative change from positive to negative .Therefore, function has local maximum at x=1.
When we goes form -2 to 0 then the sign of derivative change from negative to positive .Hence , function has local minimum at x=-1
Hence, critical points are local maximum and local minimum .