# Math Homework Help

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• Math
cot(t) = cos(t)/sin(t), csc(t) = 1/sin(t). So (cot^2(t))/csc(t) = (cos^2(t))/(sin^2(t))*(1/(1/sin(t))) =(cos^2(t))/(sin^2(t))*sin(t) = (cos^2(t))/sin(t), which is the same as the right part...

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• Math
Verify the identity: 1/tan(beta)+tan(beta)=[sec^2(beta)]/tan(beta) [1+tan^2(beta)]/tan(beta)=[sec^2(beta)]/tan(beta) Use the pythagorean identity 1+tan^2(beta)=sec^2(beta)....

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• Math
Verify the identity. sin^(1/2)(x)cos(x)-sin^(5/2)(x)cos(x)=cos^3(x)sqrt(sin(x)) Factor out the GCF sin^(1/2)(x)cos(x). sin^(1/2)(x)cos(x)[1-sin^2(x)]=cos^3(x)sqrt(sin(x)) Use the pythagorean...

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• Math
Verify the identity. sec^6(x)(sec(x)tan(x))-sec^4(x)(sec(x)tan(x))=sec^5(x)tan^3(x) Factor out the GCF sec^4(x)(sec(x)tan(x)) sec^4(x)(sec(x)tan(x))[sec^2(x)-1]=sec^5(x)tan^3(x) Use the...

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• Math
Verify cot(x)/sec(x)=csc(x)-sin(x): Working from the left side, we show that it is equivalent to the right side: cot(x)/sec(x) =((cos(x))/(sin(x)))/(1/(cos(x))) =(cos^2(x))/(sin(x))...

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• Math
Verify the identity: [sec(theta)-1]/[1-cos(theta)]=sec(theta) Use the reciprocal identity cos(theta)=1/sec(theta) [sec(theta)-1]/[1-cos(theta)]=sec(theta)...

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• Math
Verify the identity: sec(x)-cos(x)=sin(x)tan(x) The reciprocal of sec(x)=1/cos(x). sec(x)-cos(x)=sin(x)tan(x) 1/cos(x)-cos(x)=sin(x)tan(x) [1-cos^2(x)]/cos(x)=sin(x)tan(x) Use the...

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• Math
Transform the left side: sec(x)*(csc(x)-2sin(x))=(1/cos(x))*((1-2sin^2(x))/sin(x)=(cos^2(x)-sin^2(x))/(cos(x)*sin(x)). The right side is...

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• Math
By definition, sec(y)=1/cos(y). Therefore sec(y)*cos(y)=[1/cos(y)]*cos(y)=1 , QED.

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• Math
Verify the identity: cot^2(y)[sec^2(y)-1]=1 Use the pythagorean identity 1+tan^2(y)=sec^2(y). If tan^2(y) is isolated the identity would be tan^2(y)=sec^2(y)-1. cot^2(y)[sec^2(y)-1]=1...

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• Math
sin(x)=x^2-2 f(x)=x^2-2-sin(x)=0 f'(x)=2x-cos(x) See the attached graph. From the graph, the curve of the function intersects the x-axis at ~~ -1.05 and 1.70. These can be used as initial...

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• Math
f(x)=x^6-x^5-6x^4-x^2+x+10 f'(x)=6x^5-5x^4-24x^3-2x+1 x_(n+1)=x_n-((x_n)^6-(x_n)^5-6(x_n)^4-(x_n)^2+x_n+10)/(6(x_n)^5-5(x_n)^4-24(x_n)^3-2(x_n)+1) See the attached graph. From the graph the...

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• Math
x^5-3x^4+x^3-x^2-x+6=0 To solve this, using Newton's method, apply the formula:   x_(n+1)=x_n- (f(x_n))/(f'(x_n)) Let the function be:   f(x) =x^5-3x^4+x^3-x^2-x+6 Then, take its...

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• Math
x/(x^2+1)=sqrt(1-x) f(x)=x/(x^2+1)-sqrt(1-x)=0 f'(x)=((x^2+1)-x(2x))/(x^2+1)^2-(1/2)(1-x)^(-1/2)(-1) f'(x)=(1-x^2)/(x^2+1)^2+1/(2sqrt(1-x)) See the attached graph. From the graph the curve...

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• Math
cos(x^2-x) = x^4 Set the left side equal to zero. 0=x^4-cos(x^2-x) To solve using Newton's method, apply the formula: x_(n+1)=x_n - (f(x_n))/(f'(x_n)) Let the function of the given equation...

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• Math
4(e^(-x^2))sin(x)=x^2-x+1 f(x)=4(e^(-x^2))sin(x)-x^2+x-1=0 To solve equation using Newton's method apply the formula, x_(n+1)=x_n-f(x_n)/(f'(x_n))...

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• Math
You need to remember how Newton's method is used, to evaluate the roots of a transcendent equation. x_(n+1) = x_n - (f(x_n))/(f'(x_n)) , where x_n is the approximate solution to the equation...

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• Math
Approximate x=root(5)(20) by taking. f(x)=x^5-20 f'(x)=5x^4 x_(n+1)=x_n-((x_n)^5-20)/(5(x_n)^4) since root(5)(32) =2 and 32 is reasonably close to 20 , we will take x_1=2 and approximate...

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• Math
Newton's method for the equation f(x)=0 is given by x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} where x_n is the nth iteration. In this case, we need to consider the number (100)^{1/100} as...

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• Math
3cos(x)=x+1 f(x)=x+1-3cos(x)=0 To solve using Newton's method apply the formula, x_(n+1)=x_n-f(x_n)/(f'(x_n)) f'(x)=1+3sin(x) Plug in f(x) and f'(x) in the formula,...

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• Math
sqrt(x + 1) = x^2-x Set the left side equal to zero. 0=x^2-x-sqrt(x - 1) To solve using Newton's method, apply the formula: x_(n+1)=x_n-(f(x_n))/(f'(x_n)) Let the function f(x) be:...

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• Math
(x-2)^2=ln(x) f(x)=(x-2)^2-ln(x)=0 f'(x)=2(x-2)-1/x See the attached graph. The curve of the function intersects the x-axis at x ~~ 1.4 and 3. These can be used as initial approximates....

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• Math
1/x=1+x^3 Set the left side equal to zero. 0=1+x^3 -1/x To solve this using Newton's method, apply the formula: x_(n+1) = x_n - (f(x_n))/(f'(x_n)) Let the function be: f(x) = 1+x^3-1/x Take...

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• Math
x^3=tan^-1(x) f(x)=x^3-tan^-1(x)=0 f'(x)=3x^2-(1/(x^2+1)) To solve using Newton's method apply the formula, x_(n+1)=x_n-f(x_n)/(f'(x_n)) Plug in f(x) and f'(x) in the formula...

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• Math
We will solve the problem using Wolfram Mathematica (you can also use Wolfram Alpha which is free). Graph of f Plot[(x^3 + 5 x^2 + 1)/(x^4 + x^3 - x^2 + 2), {x, -10, 10}, PlotRange -> {-1,...

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• Math
We will find the derivative using Wolfram Mathematica (you can also use Wolfram Alpha which is free). For graphing we will use GeoGebra (freeware) because Mathematica cannot graph powers with...

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• Math
We will solve the problem using Wolfram Mathematica (you can also use Wolfram Alpha which is free). Graph of Plot[Sqrt[x + 5 Sin[x]], {x, -6, 20}] By writing the above line of code we obtain the...

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• Math
We will solve the problem using Wolfram Mathematica (you can also use Wolfram Alpha which is free). Graph of Plot[(x^2 - 1) E^ArcTan[x], {x, -3, 3}] By writing the above line of code we obtain the...

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• Math
We will use Wolfram Mathematica (you can also use Wolfram Alpha which is free) to solve the problem. Graph of f Plot[(1 - E^(1/x))/(1 + E^(1/x)), {x, -5, 5}] By writing the above line of...

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• Math
We will solve the problem using Wolfram Mathematica (you can also use Wolfram Alpha which is free). Graph of Plot[1/(1 + E^Tan[x]), {x, - Pi/2, Pi/2}] Plot[1/(1 + E^Tan[x]), {x, -6 Pi, 6 Pi}]...

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• Math
When c=0 , this is the basic cubic polynomial f(x)=x^3 which has no maximum, or minimum, and has horizontal tangent at x=0. Its graph is given by There is also an inflection point as the...

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• Math
We can factor the function to get f(x)=\sqrt{x^2(x^2+c)}=|x|\sqrt{x^2+c}. There are several cases. When c=0 , the domain is all real numbers, and the function becomes the simple...

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• Math
This function is defined only on [-c,c] (we can assume that c is positive) and infinitely differentiable inside the interval. f'_c(x) = sqrt(c^2-x^2)+x*(-2x)/(2sqrt(c^2-x^2)) =...

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• Math
To find extremums and inflection points, first find f' and f'': f'_c(x) = e^x-ce^(-x), f''_c(x) = e^x+ce^(-x). 1. For c<0:f' is always positive, f increases.f'' has one root,...

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• Math
For c>0 f(x) is defined and infinitely differentiable everywhere. For c=0 f is undefined at x=0, and for c<0 f is undefined for x<=sqrt(-c) and x>+-sqrt(-c). Find f' an f'': f'_c(x) =...

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• Math
f_c(x) is defined everywhere and is infinitely differentiable. f_c(x) = g(cx) where g(x) = x/(1+x^2). So f'_c(x) = g'(cx)*c = c*(1-(cx)^2)/(1+(cx)^2)^2, f''_c(x) = c^2*g''(cx) =...

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• Math
Although this question looks tough, it is really simple if you break it down into familiar component parts. f(x) is composed of the component "basis" functions which include x^2 and e^-x. I would...

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• Math
f(x)=x^6-15x^5+75x^4-125x^3-x f'(x)=6x^5-75x^4+300x^3-375x^2-1 f''(x)=30x^4-300x^3+900x^2-750x See the attached graph and link. f is in red color , f' is in blue color and f'' is in green...

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• Math
See the attached graph of f(image) in red color and links.Graph is drawn on two ranges for better clarity. From the graph , f is decreasing on intervals (-oo ,-15) and (4.40,19.0) f is increasing...

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• Math
The denominator of the function vanishes at x\approx -1.465571 which means there is a vertical asymptote there. Since the numerator is of degree one and the denominator is degree three, there is...

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• Math
f(x)=6sin(x)-x^2 Domain -5<=x<=3 f'(x)=6cos(x)-2x f''(x)=-6sin(x)-2 Refer the graphs plotted in several ranges for clarity (attached image and links). f(x) plotted in red color,f'(x)...

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• Math
f(x)=6sin(x)+cot(x) See the attached graph and link(different range), f(x) is in Red color, f' is in Blue color and f'' is in Green color. From the graph, Vertical Asymptotes at x=0, x=pi ,...

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• Math
f(x)=e^x-0.186x^4 f'(x)=e^x-0.186(4x^3) f'(x)=e^x-0.744x^3 f''(x)=e^x-0.744(3x^2) f''(x)=e^x-2.232x^2 Refer the attached graph in image and link. f(x) plotted in red color,f'(x) in blue...

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• Math
See the attached graph and links. Graph is drawn in several ranges to have clarity. From the graph, f is decreasing in the intervals about (- oo , -16),(-0.2 ,0) and (0,oo ) f is increasing in...

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• Math
Let c=2\cdot 10^8 . Then rearranging the function, we get f(x)=1/x^8-c/x^4 = x^{-8}-cx^{-4}=\frac{1-cx^4}{x^8} . The derivative is found using the power rule to get f'(x)=-8x^{-9}+4cx^{-5}...

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• Math
f(x)=((x+4)(x-3)^2)/(x^4(x-1)) Vertical asymptotes are real zeros of the denominator of the function. x^4(x-1)=0 x=0 , x=1 Vertical asymptotes are at x=0 and x=1 Degree of numerator =3...

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• Math
f(x)=((2x+3)^2(x-2)^5)/(x^3(x-5)^2) Vertical asymptotes are the undefined points, also known as zeros of denominator. Let's find the zeros of denominator of the function, x^3(x-5)^2=0 x^3=0...

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• Math
See the attached graphs and links. f(x) is in red color , f'(x) in blue color and f''(x) in green color.Graphs are plotted in various ranges to have clarity. From the graphs f(x)=0 hArr x=0.5 ,...

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• Math
sec^2(x)-4sec(x)=sec(x)*(sec(x)-4)=0. So sec(x)=0 or sec(x)=4. sec(x)=1/cos(x) and it is never zero. It is =4 when cos(x)=1/4. There are two solutions on (0, 2pi): arccos(1/4) and 2pi-arccos(1/4)....

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Denote sec(x) as y and obtain a quadratic equation for y: y^2+2y-8=0. This equation has two roots, y_1=2 and y_2=-4. So we have to solve sec(x)=2 and sec(x)=-4 separately. sec(x) = 1/cos(x),...