Template question. The student is asked to perform a two factor ANOVA to test the null hypotheses that the measurement does not depend on each of the factors, and that there is no interaction between the factors.

Solving a system of three linear equations in 3 unknowns using Gauss Elimination in 4 stages. Solutions are all integral.

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$I$ compact interval, $g:I\rightarrow I,\;g(x)=ax^3+bx^2+cx+d$. Find stationary points, local and global maxima and minima of $g$ on $I$

$I$ compact interval, $g:I\rightarrow I$, $g(x)=(x-a)(x-b)^2$. Stationary points in interval. Find local and global maxima and minima of $g$ on $I$. 

$g: \mathbb{R} \rightarrow \mathbb{R}, g(x)=\frac{ax}{x^2+b^2}$. Find stationary points and local maxima, minima. Using limits, has $g$ a global max, min? 

$I$ compact interval. $\displaystyle g: I\rightarrow I, g(x)=\frac{x^2}{(x-c)^{a/b}}$. Are there stationary points and local maxima, minima? Has $g$ a global max, global min? 

$I$ compact interval. $\displaystyle g: I \rightarrow I, g(x)=\frac{ax}{x^2+b^2}$. Find stationary points and local maxima, minima. Using limits, has $g$ a global max, min? 

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Template question. The student is asked to perform a two factor ANOVA to test the null hypotheses that the measurement does not depend on each of the factors, and that there is no interaction between the factors.

Solve for $x$: $\displaystyle 2\log_{a}(x+b)- \log_{a}(x+c)=d$. 

Make sure that your choice is a solution by substituting back into the equation.

Solve for $x$:  $\log_{a}(x+b)- \log_{a}(x+c)=d$

How to input x^2 and x*y.

Find the $x$ and $y$ components of a force which is applied at an angle to a particle. Resolve using $F \cos \theta$. The force acts in the positive $x$ and positive $y$ direction.

Find the $x$ and $y$ components of a force which is applied at an angle to a particle. Resolve using $F \cos \theta$. The force is applied in the negative $x$ and negative $y$ direction.

Find the $x$ and $y$ components of a force which is applied at an angle to a particle. Resolve using $F \cos \theta$. The force is applied in the negative $x$ direction but the positive $y$.