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4

Let $f(x)=x^4-3x^2+x-\sin x$. By inspection $f$ has a root at $0$, and we see that $f(-1)<0$, $f(1)<0$, $f(2)>0$, $f(-2)>0$, so, applying the Intermediate Value Theorem, $f$ has at least two other roots. Consider now $f'(x)=4x^3-6x+1-\cos x$. Since $f'(0)=0$, $f$ has a double root at $0$, and so, counting multiplicities, $f$ has at least 4 ...


3

Divide both sides by $6^x$: $$\tag1\left(\frac12\right)^x+\left(\frac23\right)^x+\left(\frac56\right)^x=1.$$ Note that $f(x)=q^x$ with $0<q<1$ has the property that $f(0)=1$, $f$ is continuous and strictly decreasing on $[0,\infty)$ and $f(x)\to 0$ as $x\to\infty$. Therefore the left hand side in $(1)$ is $3$ at $x=0$, is continuous and structliy ...


2

Quick: $$\lim_{x\to\pm\infty}\cos x\,\;or\;\,\sin x\;\;\;\;\;\text{don't exist}$$ and this already takes off options (1) and (3). In (4) we have a function that goes to zero ($\;x\;$) times a bounded function $\;\left(\sin\frac1x\right)\;$ and thus the limit is zero, yet it doesn't equal the defined value of the function there thus (4) not true.


2

First question, yes. It is a vertical translation since it takes it multiplies the output of $f(x)$ by $8$. I'm not entirely clear what you are asking in the second question, but note: $$f(x) = x^2 \implies f(x/\sqrt 8) = \left(\frac x{\sqrt 8}\right)^2 = \dfrac {x^2}{8} \neq 8x^2$$ That flattens (vertically) the graph of $f(x) = x^2$ by a factor of ...


2

In physics, this is usually used as dependent versus independent as in a velocity versus time or position versus time graphs. See Motion graphs and derivatives as well as from Line chart we have "The chart can then be referred to as a graph of 'Quantity one versus quantity two, plotting quantity one up the y-axis and quantity two along the x-axis.' " ...


2

A function is a relation between a set of inputs and a set of permissible outputs (determined by the relation), with the property that each input is related to exactly one output. So you need to make sure to change "only" to "exactly" or to " one and only one". I emphasize one and only one because each input value must be related to one output value, and ...


1

Start with the graph of $y=f(x)=x^{3}$. Then draw $g(x)=(x-1)^{3}=f(x-1)$. This graph can be obtained by translating the graph of $f(x)$ to the right by +1 (Why?). Then draw $h(x)=\frac{1}{2}(x-1)^{3}=\frac{1}{2}g(x)$. Note that the values of $h(x)$ are reduced by a factor of $1/2$ with respect to the equivalent ones of $g(x)$. Finally draw the graph ...


1

A function $f:X \to Y$ is a subset $A\subset X × Y$ subject to the condition that every element of $X$ is the first component of one and only one ordered pair in the subset $A$. Hence, the picture depicts a one-to-many mapping - this is not a function.


1

Transforming $f(x)=x^2$ into $g(x)=8x^2$ can both be interpreted as a vertical stretch by a factor of $8$, or it can also be interpreted as a horizontal compression, by a factor of $\sqrt{8}$: $8f(x)=8\cdot x^2=8x^2 \;\;(\text{This is a vertical stretch by factor of 8})$ $f(\sqrt{8}x)=((\sqrt{8}x))^2=8x^2\;\; (\text{This is a horizontal compression by a ...


1

Here is my approach to finding this result. I've just started using Geoalgebra so this was a nice problem to try things in it on, here's the diagram I made for this problem (though it would be identical to hand drawing it in the space the question provides, but for me at least the computer is a lot neater!) Firstly in an attempt to make this problem ...


1

Suppose the root is $$a\pm bi$$ Then the polynomial will have a factor of the form $$(x-a-bi)(x-a+bi)=(x-a)^2-(bi)^2=(x-a)^2+b^2$$ So we will have for some polynomial $g$, $$f(x)=(x-a)^2g(x)+b^2g(x)$$ Thus $$f'(x)=2(x-a)g(x)+\big((x-a)^2+b^2\big)\,g'(x)$$ and $$f'(a)=b^2\,g'(a)$$ If f is quadratic, then g will be constant, so we will get $$f'(a)=0$$ In other ...


1

Setting multiplicities aside, one approach would be to divide by $x$ and graph the even function $$f(x)={\sin x\over x}-1$$ (noting that $f(0)=0$) and the odd function $$g(x)=x^3-3x$$ Even a rough sketch shows the two curves cross transversally three times: once at $x=0$, once in the interval $[1,2]$, and once in the interval $[-2,-1]$, for a total of ...


1

Let's first distribute same labels: $\hskip1.5in$ Let the $C$ lie at $(1,0)$, $F$ at $a(1,0)$, $A$ at $(\cos(2\pi/3),-\sin(2\pi/3))$ and $D$ at $a(\cos(2\pi/3),-\sin(2\pi/3))$. You need to check whether $|CD|=|AF|$ and this is: $$ |(1-a\cos(2\pi/3),-a\sin(2\pi/3))|=|(a-\cos(2\pi/3),-\sin(2\pi/3))|\\ ...


1

Try this: set( gca, 'Position', get( gca, 'OuterPosition' ) - ... get( gca, 'TightInset' ) * [-1 0 1 0; 0 -1 0 1; 0 0 1 0; 0 0 0 1] ); Here is an example of what the above does: figure plot( 1:10 ) ylabel( 'Y Axis' ); xlabel( 'X Axis' ); Before: After: And side-by-side before the printing:


1

That's correct: it's not always equal to zero. Nor should you expect it to be. Let's look at a simpler example, in one variable, $x$. Let $$f(x)=x$$ The equation $f(x)=x=0$ represents a set of solutions: in this case, the set $\{0\}$. Differentiating, we get $f'(x)=1=0$, which is clearly absurd, and should indicate that you can't do such a thing and expect ...


1

Draw the following lines: $2x+3, -\frac{1}{2}x+2,3x-5$ From their graphs find their y-intercepts, and slopes. Try to notice a relationship between the slopes, intercept and the equation of the line Try to explain this relationship (prove it) If you find difficulties explaining/proving this relationship, please make another post about what you attempted ...


1

If you're finding this problem tricky it might be easier to draw a small sketch. If you have no squared paper you can draw eight or ten squares onto blank paper yourself. If the line crosses the y axis at -3/4 you already have the equation for x = 0, your first point, and if you use the squares to carefully draw a short section of slope with the angle ...


1

Here's the Mathematica code: RegionPlot3D[ 0 <= x && x <= 1 && x^2 <= y && y <= x && 0 <= z && z <= x, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, AxesLabel -> {x, y, z}, PlotPoints -> 35 ] and the output:


1

In Mathematica you get a plot of the region using RegionPlot3D[0 <= x <= 1 && x^2 <= y <= x && 0 <= z <= x, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}] The first argument you'll easily identify as your region specification. The following arguments give the variables and the range which should be shown (you generally ...


1

In Maple this can be done with plots:-implicitplot3d, but a sharper result is obtained more easily by simply using the plot3d command with its filled option. The upper z-surface is simply z=x, and the lower surface is z=0, and filling between them corresponds to your constraints z>=0 and z<=x. plot3d(x, y = x .. x^2, x = 0 .. 1, ...



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