# test

T-chart

Graph

$$\displaystyle y=-\frac{1}{2}\sqrt{{x+2}}+4$$

Parent function:

$$y=\sqrt{x}$$

For square root functions, use 0, 1, and 4 for the $$x$$ values of the parent function.

 x – 2 x y $$-\frac{1}{2}y+4$$ –2 0 0 4 –1 1 1 3.5 2 4 2 3

Domain: $$\left[ {-2,\infty } \right)$$

Range:  $$\left( {3,\infty } \right]$$

Vertical compression by factor of $$\frac{1}{2}$$ (over 1, down $$\frac{1}{2}$$), reflect over the $$x$$-axis, translate 2 units right, 4 units up.

$$y=\sqrt[3]{{-2\left( {x-1} \right)}}+2$$

Parent function:

$$y=\sqrt[3]{x}$$

For cube root functions, use –1, 0, and 1 for the $$x$$ values of the parent function.

 $$-\frac{1}{2}x+1$$ x y y + 2 $$\displaystyle 1\frac{1}{2}$$ –1 –1 1 1 0 0 2 $$\frac{1}{2}$$ 1 1 3

Domain:  $$\left( {-\infty ,\infty } \right)$$

Range: $$\left( {-\infty ,\infty } \right)$$

Horizontal compression by a factor of $$\frac{1}{2}$$ (over $$\frac{1}{2}$$, down 2), reflect over the $$y$$-axis, translate 1 unit right, 2 units up.

**Notes on End Behavior: To get the end behavior of a function, we just look at the smallest and largest values of $$x$$, and see which way the $$y$$ is going. Not all functions have end behavior defined; for example, those that go back and forth with the $$y$$ values and never really go way up or way down (called “periodic functions”) don’t have end behaviors.

Most of the time, our end behavior looks something like this:$$\displaystyle \begin{array}{l}x\to -\infty \text{, }\,y\to \,\,?\\x\to \infty \text{, }\,\,\,y\to \,\,?\end{array}$$ and we have to fill in the $$y$$ part. For example, the end behavior for a line with a positive slope is: $$\begin{array}{l}x\to -\infty \text{, }\,y\to -\infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}$$, and the end behavior for a line with a negative slope is: $$\begin{array}{l}x\to -\infty \text{, }\,y\to \infty \\x\to \infty \text{, }\,\,\,y\to -\infty \end{array}$$. One way to think of end behavior is that for $$\displaystyle x\to -\infty$$, we look at what’s going on with the $$y$$ on the left-hand side of the graph, and for $$\displaystyle x\to \infty$$, we look at what’s happening with $$y$$ on the right-hand side of the graph.

There are a couple of exceptions; for example, sometimes the $$x$$ starts at 0 (such as in the radical function), we don’t have the negative portion of the $$x$$ end behavior. Also, when $$x$$ starts very close to 0 (such as in in the log function), we indicate that $$x$$ is starting from the positive (right) side of 0 (and the $$y$$ is going down); we indicate this by $$\displaystyle x\to {{0}^{+}}\text{, }\,y\to -\infty$$.

 Systems of Equations Calculator Steps Notes We would get a situation like this if we had the two equations below. Notice that the slope of the two equations are the same, but the $$y$$-intercepts are different; thus the lines are parallel:     These types of equations are called inconsistent, since there are no solutions.   Notice that if we were to “solve” the two equations, we’d end up with “6 = 8”, and no matter what $$d$$ or $$j$$ is, 6 can never equal 8. Thus, there are no solutions. Sometimes we have a situation where the system contains the same equations even though it may not be obvious:   See how we may not know unless we actually graph, or simplify them?   These types of equations are called dependent or coincident since they are one and the same equation and they have an infinite number of solutions. Since they have at least one solution, they are also consistent.   Notice that if we were to “solve” the two equations, we’d end up with “6 = 6”, and no matter what $$d$$ or $$j$$ is, 6 always equals 6. Thus, there are an infinite number of solutions, but $$d$$ always has to be equal to –j + 6. We can write the solution as ( j,  –j + 6).

 Curve Sketching Problem:   The graph to the right is the derivative of a function f. a)      The graph f  has a local maximum at x = ? b)      The graph f  has a local minimum at x = ? The graph f  has a point of inflection (POI) at x = ? Solution:   Given the derivative, let’s first try to draw the function.  Remember P→M→S – if we go backwards from the derivative to the original function, where there are minimums/maximums in the derivative, there are POI in the original, and where there are sign changes in the derivative, there are minimums/maximums in the original. Now let’s answer the questions from above: a)      When a function has a local maximum, its derivative has a sign change from positive to negative, so there is a local maximum at x = 4. b)      When a function has a local minimum, its derivative has a sign change from negative to positive, so there are no local minimums.  (Note: in some textbooks, endpoints can be relative extrema, so we’d have local minimums at x = 0 and x = 5). When a function has a POI, its derivative has a minimum or maximum, so there are POIs at x = 1 and x = 3.

 Curve Sketching Problem: Find all relative extrema, with $$y$$ values, and points of inflection for $$f\left( x \right)=6{{x}^{{\frac{2}{3}}}}-x$$. Find where the function is increasing and decreasing, and where it is concave up and down. Solution: First take the first and second derivatives: $$\displaystyle f\left( x \right)=6{{x}^{{\frac{2}{3}}}}-x;\,\,\,\,\,\,{{f}^{\prime}(x)}=4{{x}^{{-\frac{1}{3}}}}-1;\,\,\,\,\,\,\,{{f}^{\prime \prime}(x)}=-\frac{4}{3}{{x}^{{-\frac{4}{3}}}}$$. Next, find the increasing (decreasing) intervals: where the derivative is positive (negative). Set the first derivative to 0 first to find critical point(s) and use test points with the derivative to find increasing/decreasing intervals: $$\displaystyle 4{{x}^{{-\frac{1}{3}}}}-1=0;\,\,\,\,\,{{\left( {4{{x}^{{-\frac{1}{3}}}}} \right)}^{{-3}}}={{\left( 1 \right)}^{{-3}}};\,\,\,\frac{1}{{64}}x=1;\,\,\,\,x=64;\,\,\,\text{also undefined at }x=0$$:    .  The critical points are at $$x=0$$ and $$x=64$$. To get the $$y$$ points, plug in $$x$$ in the original equation: $$\displaystyle y=6{{\left( 0 \right)}^{{\frac{2}{3}}}}-0=0;\,\,\,\,y=6{{\left( {64} \right)}^{{\frac{2}{3}}}}-64=32$$. The critical points are $$\left( {0,0} \right)$$ (min) and $$\left( {64,32} \right)$$ (max).   Set the second derivative to 0 to find point of inflections (POI) and intervals of concavity, using test points with the second derivative: $$\displaystyle \frac{4}{3}{{x}^{{-\frac{4}{3}}}}=0;\,\,\,\,\,{{\left( {\frac{4}{3}{{x}^{{-\frac{4}{3}}}}} \right)}^{{-\frac{3}{4}}}}={{0}^{{-\frac{3}{4}}}};\,\,\,\text{undefined at 0}$$: . Since both intervals are negative, there is not a POI.   We have the following characteristics; notice how we have to “jump” over (exclude) $$x=0$$ in some cases:   Increasing Intervals: $$\left( {0,64} \right)$$         Decreasing Intervals: $$\left( {-\infty ,0} \right)\cup \left( {64,\infty } \right)$$   Local Minimum:  $$\left( {0,0} \right)$$         Local Maximum: $$\left( {64,32} \right)$$ Concave Up Intervals: none          Concave Down Intervals:  $$\left( {-\infty ,0} \right)\cup \left( {0,\infty } \right)$$           Points of Inflection (POI): none

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 Second Derivative Problem and Graph Solution Using the Second Derivative Test, find relative extrema for the following function, if it’s possible to do so: is a relative minimum, and  is a relative maximum. Using the Second Derivative Test, find relative extrema for the following function, if it’s possible to do so: Note that the Second Derivative Test does not apply, since the first derivative fails to exist (undefined) at  (there are actually vertical asymptotes at those values).  So we don’t have to take the second derivative. We also see that the first derivative will always be positive, so there are no relative extrema (maximums or minimums).
 Second Derivative Problem and Graph Solution Find the point of inflection and discuss the concavity of the function:   $$f\left( x \right)={{x}^{3}}-5{{x}^{2}}+8x$$ \begin{align}f\left( x \right)&={{x}^{3}}-5{{x}^{2}}+8x\\{f}’\left( x \right)&=3{{x}^{2}}-10x+8;\,\,\,\text{Critical values:}\,\,\,\left( {3x-4} \right)\left( {x-2} \right)=0;\,\,\,x=\frac{4}{3},\,\,2\\{f}^{\prime \prime}\left( x \right)&=6x-10;\,\,\,\text{Point of Inflection (POI):}\,\,\,6x-10=0;\,\,\,x=\frac{5}{3}\end{align}   Create sign chart by testing intervals for both first and second derivatives:                                                       The point of inflection, where the graph changes from “cup down” (concave down) to “cup up” (concave up) is at $$\displaystyle x=\frac{5}{3}$$. We see that the graph is concave down in the interval $$\displaystyle \left( {-\infty ,\frac{5}{3}} \right)$$ and is concave up in the interval $$\displaystyle \left( {\frac{5}{3},\infty } \right)$$. We also see that the graph is increasing in the intervals $$\displaystyle \left( {-\infty ,\frac{4}{3}} \right)$$ and $$\left( {2,\infty } \right)$$ and decreasing in the interval $$\displaystyle \left( {\frac{4}{3},2} \right)$$. Note that $$\displaystyle \left( {\frac{4}{3},\,{{{\left( {\frac{4}{3}} \right)}}^{3}}-5\cdot {{{\left( {\frac{4}{3}} \right)}}^{2}}+8\cdot \frac{4}{3}} \right)=\left( {\frac{4}{3},\frac{{112}}{{27}}} \right)$$ is a relative maximum and $$\left( {2,\,{{2}^{3}}-5\cdot {{2}^{2}}+8\cdot 2} \right)=\left( {2,4} \right)$$ is a relative minimum. Find the point of inflection and discuss the concavity of the function:   $$\displaystyle f\left( x \right)=\frac{2}{{{{x}^{2}}+1}}$$ \displaystyle \begin{align}f\left( x \right)&=\frac{2}{{{{x}^{2}}+1}}=2{{\left( {{{x}^{2}}+1} \right)}^{{-1}}}\\{f}’\left( x \right)&=-2{{\left( {{{x}^{2}}+1} \right)}^{{-2}}}\cdot 2x=-4x{{\left( {{{x}^{2}}+1} \right)}^{{-2}}}=\frac{{-4x}}{{{{{\left( {{{x}^{2}}+1} \right)}}^{2}}}};\,\,\,\text{Critical value :}\,\,\,x=0\\{f}^{\prime \prime}\left( x \right)&=\frac{{{{{\left( {{{x}^{2}}+1} \right)}}^{2}}\left( {-4} \right)-\left( {-4x} \right)\left[ {2\left( {{{x}^{2}}+1} \right)\left( {2x} \right)} \right]}}{{{{{\left( {{{x}^{2}}+1} \right)}}^{4}}}}=\frac{{4\left( {3{{x}^{2}}-1} \right)}}{{{{{\left( {{{x}^{2}}+1} \right)}}^{3}}}}\text{;}\,\,\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\text{POI: }\,\,3{{x}^{2}}-1=0;\,\,\,x=\pm \frac{{\sqrt{3}}}{3}\text{ }\end{align} By checking points in intervals, the graph is concave up in the intervals $$\displaystyle \left( {-\infty ,-\,\frac{{\sqrt{3}}}{3}} \right)$$ and $$\displaystyle \left( {\frac{{\sqrt{3}}}{3},\infty } \right)$$, and is concave down in the interval $$\displaystyle \left( {-\frac{{\sqrt{3}}}{3},\frac{{\sqrt{3}}}{3}} \right)$$, with a relative maximum at $$\left( {0,2} \right)$$.
 Problem Solution

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Vector ProblemSolution

Find x so that the following vectors are orthogonal:

(a)

(b)

We’ll need to find  and make sure it equals 0:

(a)

(b)

 Vector Parametric Problem Solution Find the equation of the plane that passes through the point $$\left( {4,0,-1} \right)$$ and is perpendicular to the vector $$n=\text{i}-3\text{j}+2\text{k}$$. The equation of this plane is $$\displaystyle x-3y+2z=2$$. Find the equation of the plane that passes through the point $$\left( {3,-3,1} \right)$$, and is perpendicular to the line $$\displaystyle \frac{{2-x}}{3}=\frac{{y+1}}{4}=\frac{z}{2}$$. We know the direction vector of this line is $$\left\langle {-3} \right.,\left. {4,2} \right\rangle$$ (we have to set each expression to $$t$$, and solve back for$$x$$, $$y$$, and $$z$$. Also note that we had to multiply the $$x$$ expression by –1).         The equation of this plane is $$\displaystyle -3x+4y+2z=-19$$.

 Quadratic with Complex Solutions Solve Using Quadratic Formula $${{x}^{2}}-2x+2=0$$   $$\begin{array}{l}a=1\\b=-2\\c=2\end{array}$$ \require{cancel} \displaystyle \begin{align}\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}&=\frac{{-\left( {-2} \right)\pm \sqrt{{{{{\left( {-2} \right)}}^{2}}-4\left( 1 \right)\left( 2 \right)}}}}{{2\left( 1 \right)}}\\&=\frac{{2\pm \sqrt{{-4}}}}{2}=\frac{{2\pm 2i}}{2}\\&=\frac{{\cancel{2}(1\pm i)}}{{\cancel{2}}}=1\pm i\end{align} $$3{{x}^{2}}-2x+2=0$$   $$\begin{array}{l}a=3\\b=-2\\c=2\end{array}$$ \displaystyle \begin{align}\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}&=\frac{{-\left( {-2} \right)\pm \sqrt{{{{{\left( {-2} \right)}}^{2}}-4\left( 3 \right)\left( 2 \right)}}}}{{2\left( 3 \right)}}\\&=\frac{{2\pm \sqrt{{-20}}}}{6}=\frac{{2\pm 2\sqrt{5}\,i}}{6}\\&=\frac{{{}^{1}\cancel{2}(1\pm \sqrt{5}\,i)}}{{{{{\cancel{6}}}^{3}}}}=\frac{1}{3}\pm \frac{{\sqrt{5}i}}{3}\text{ }\text{ or }\text{ }\frac{1}{3}\pm \frac{{\sqrt{5}}}{3}i\text{ }\end{align}

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