# Consider this multivariable function. f(x,y)=xy+2x+y−36 a) What is the value of f(2,−3)? b) Find all x-values such that f (x,x) = 0

Question
Multivariable functions
Consider this multivariable function. f(x,y)=xy+2x+y−36
a) What is the value of f(2,−3)?
b) Find all x-values such that f (x,x) = 0

2021-02-10
a) We find f(2,-3) be replacing x by 2 and by -3 in f(x,y)
f(2,-3)=2(-3)+2(2)+(-3)-36
f(2,-3)=-6+4-3-36=-41
b) Now, we solve f(x,x)=0 as follow.
f(x,x)=0
$$\displaystyle{x}^{{2}}+{2}{x}+{x}-{36}={0}$$
$$\displaystyle{x}^{{2}}+{3}{x}-{36}={0}$$
$$\displaystyle{x}=\frac{{-{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}={\left(-{3}\pm\sqrt{{{3}^{{2}}-{4}{\left({1}\right)}{\left(-{36}\right)}}}\right)}\frac{{)}}{{{2}{\left({1}\right)}}}=\frac{{-{3}\pm\sqrt{{153}}}}{{2}}$$

### Relevant Questions

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Previous studies show that $$\sigma_1 = 19$$.
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The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
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At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
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Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
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