In each case, for a) f has local maximum at (1,1) and for b) f has saddle point at (1,1).
a) f_{xx}f_{yy}-(f_{xy})2
=(-4)(-2)-(1)2
=8-1
=7>0
f_{xx}=-4<0
Therefore, f has local maximum at (1,1)
b) f_{xx}f_{yy}-(f_{xy})2
=(-4)(-2)-(3)2
=8-9
=-1<0
Therefore, f has saddle point at (1,1)
A factor at which a feature of variables has partial derivatives identical to 0 however at which the feature has neither a most nor a minimal value.
To learn more about derivatives check the link below:
brainly.com/question/28376218
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Complete question:
Suppose (1, 1) is a critical point of a function f with continuous second derivatives. In each case, what can you say about f?a) f_{xx}f_{yy}-(f_{xy})2 and b) f_{xx}f_{yy}-(f_{xy})2
88/80=1.1
1.1X100=110
the answer is 110 $ was the regular price
25 / 32 ÷ ( -5/6)
=(25/32) (6/−5)
=(25)(6) / (32)(−5)
=150 / −160
=−15/16
(Decimal: -0.9375)
Answer:
4.75% probability that the battery will last more than 9 hours before running out of power
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
![\mu = 8, \sigma = 0.6](https://tex.z-dn.net/?f=%5Cmu%20%3D%208%2C%20%5Csigma%20%3D%200.6)
What is the probability that the battery will last more than 9 hours before running out of power g
This is 1 subtracted by the pvalue of Z when X = 9. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{9 - 8}{0.6}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B9%20-%208%7D%7B0.6%7D)
![Z = 1.67](https://tex.z-dn.net/?f=Z%20%3D%201.67)
has a pvalue of 0.9525
1 - 0.9525 = 0.0475
4.75% probability that the battery will last more than 9 hours before running out of power