2sin2(x)+5cos(x)=42sin2(x)+5cos(x)=4
- MATH
20−−√5205 Use Newton’s method to approximate the given number correct to eight decimal…
Approximate x=20−−√5x=205 by taking. f(x)=x5−20f(x)=x5-20 f'(x)=5x4f′(x)=5×4 xn+1=xn−(xn)5−205(xn)4xn+1=xn-(xn)5-205(xn)4 since 32−−√5325 =2 and 32 is reasonably close to 20 , we will take x_1=2 and approximate…
1 educator answer
- MATH
f(x)=cx+sin(x)f(x)=cx+sin(x) Describe how the graph of ffvaries as cc varies. Graph several…
To investigate what happens to maximum and minimum points we will take a look at the first derivative. f'(x)=c+cosxf′(x)=c+cosx Maximum and minimum points are where f'(x)=0.f′(x)=0. Hence we get equation
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- MATH
f(x)=x2+ce−xf(x)=x2+ce-x Describe how the graph of ffvaries as cc varies. Graph several…
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)=cx1+(c2)(x2)f(x)=cx1+(c2)(x2) Describe how the graph of ffvaries as ccvaries. Graph…
fc(x)fc(x) is defined everywhere and is infinitely differentiable. fc(x)=g(cx)fc(x)=g(cx) where g(x)=x1+x2.g(x)=x1+x2. So f’c(x)=g'(cx)⋅c=c⋅1−(cx)2(1+(cx)2)2,f′c(x)=g′(cx)⋅c=c⋅1-(cx)2(1+(cx)2)2,
2 educator answers
- MATH
f(x)=ln(x2+c)f(x)=ln(x2+c) Describe how the graph of ffvaries as cc varies. Graph several…
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”:
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- MATH
f(x)=ex+ce−xf(x)=ex+ce-x Describe how the graph of ffvaries as cc varies. Graph several…
To find extremums and inflection points, first find f’ and f”: f’c(x)=ex−ce−x,f′c(x)=ex-ce-x, f”c(x)=ex+ce−x.f′′c(x)=ex+ce-x. 1. For c<0:f’ is always positive, f increases.f” has one root,…
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- MATH
f(x)=xc2−x2−−−−−−√f(x)=xc2-x2 Describe how the graph of ffvaries as cc varies. Graph several…
This function is defined only on [-c,c] (we can assume that c is positive) and infinitely differentiable inside the interval.
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- MATH
f(x)=x4+cx2−−−−−−−√f(x)=x4+cx2 Describe how the graph of ffvaries as cc varies. Graph several…
We can factor the function to get f(x)=x2(x2+c)−−−−−−−−−√=|x|x2+c−−−−−√f(x)=x2(x2+c)=|x|x2+c. There are several cases. When c=0c=0 , the domain is all real numbers, and the function becomes the simple…
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- MATH
f(x)=x3+cxf(x)=x3+cx Describe how the graph of ffvaries as cc varies. Graph several members…
When c=0c=0 , this is the basic cubic polynomial f(x)=x3f(x)=x3 which has no maximum, or minimum, and has horizontal tangent at x=0x=0. Its graph is given by There is also an inflection point as the…
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- MATH
f(x)=11+etan(x)f(x)=11+etan(x) Use a computer algebra system to graph ff and to find f’f′
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
f(x)=1−e1x1+e1xf(x)=1-e1x1+e1x Use a computer algebra system to graph ff and to find…
We will use Wolfram Mathematica (you can also use Wolfram Alpha which is free) to solve the problem. Graph of ff 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[(x^2 – 1) E^ArcTan[x], {x, -3, 3}] By writing the above line of code we obtain the…
1 educator answer
- 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…
1 educator answer
- MATH
f(x)=x231+x+x4f(x)=x231+x+x4 Use a computer algebra system to graph ff and to find f’f′
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
f(x)=x3+5×2+1×4+x3−x2+2f(x)=x3+5×2+1×4+x3-x2+2 Use a computer algebra system to graph ff…
We will solve the problem using Wolfram Mathematica (you can also use Wolfram Alpha which is free). Graph of ff Plot[(x^3 + 5 x^2 + 1)/(x^4 + x^3 – x^2 + 2), {x, -10, 10}, PlotRange -> {-1,…
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- MATH
f(x)=(2x+3)2(x−2)5×3(x−5)2f(x)=(2x+3)2(x-2)5×3(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, x3(x−5)2=0x3(x-5)2=0
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- MATH
f(x)=(x+4)(x−3)2(x4)(x−1)f(x)=(x+4)(x-3)2(x4)(x-1) Sketch the graph by hand using asymptotes and…
f(x)=(x+4)(x−3)2×4(x−1)f(x)=(x+4)(x-3)2×4(x-1) Vertical asymptotes are real zeros of the denominator of the function. x4(x−1)=0x4(x-1)=0 x=0,x=1x=0,x=1 Vertical asymptotes are at x=0 and x=1 Degree of numerator =3…
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- MATH
f(x)=1×8−2⋅108x4f(x)=1×8-2⋅108×4 Produce graphs of ff that reveal all the important…
Let c=2⋅108c=2⋅108 . Then rearranging the function, we get f(x)=1×8−cx4=x−8−cx−4=1−cx4x8f(x)=1×8-cx4=x-8-cx-4=1-cx4x8 . The derivative is found using the power rule to get f'(x)=−8x−9+4cx−5f′(x)=-8x-9+4cx-5…
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- MATH
f(x)=1+1x+8×2+1x3f(x)=1+1x+8×2+1×3 Produce graphs of ff that reveal all the important…
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 (- ∞∞ , -16),(-0.2 ,0) and (0,∞∞ ) f is increasing in…
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- MATH
f(x)=ex−0.186x4f(x)=ex-0.186×4 Produce graphs of ff that reveal all the important aspects of the…
f(x)=ex−0.186x4f(x)=ex-0.186×4 f'(x)=ex−0.186(4×3)f′(x)=ex-0.186(4×3) f'(x)=ex−0.744x3f′(x)=ex-0.744×3 f”(x)=ex−0.744(3×2)f′′(x)=ex-0.744(3×2) f”(x)=ex−2.232x2f′′(x)=ex-2.232×2 Refer the attached graph in image and link. f(x) plotted in red color,f'(x) in blue…
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- MATH
f(x)=6sin(x)+cot(x),−π≤x≤πf(x)=6sin(x)+cot(x),-π≤x≤π Produce graphs of ff that reveal all the…
f(x)=6sin(x)+cot(x)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)=6sin(x)−x2,−5≤x≤3f(x)=6sin(x)-x2,-5≤x≤3 Produce graphs of ff that reveal all the important…
f(x)=6sin(x)−x2f(x)=6sin(x)-x2 Domain −5≤x≤3-5≤x≤3 f'(x)=6cos(x)−2xf′(x)=6cos(x)-2xf”(x)=−6sin(x)−2f′′(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)=xx3+x2+1f(x)=xx3+x2+1 Produce graphs of ff that reveal all the important aspects of…
The denominator of the function vanishes at x≈−1.465571x≈-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)=x2−140×3+x+1f(x)=x2-140×3+x+1 Produce graphs of ff that reveal all the important…
f(x)=x2−140×3+x+1f(x)=x2-140×3+x+1 f'(x)=(40×3+x+1)(2x)−(x2−1)(120×2+1)(40×3+x+1)2f′(x)=(40×3+x+1)(2x)-(x2-1)(120×2+1)(40×3+x+1)2 f'(x)=(80×4+2×2+2x)−(120×4+x2−120×2−1)(40×3+x+1)2f′(x)=(80×4+2×2+2x)-(120×4+x2-120×2-1)(40×3+x+1)2…
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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 (-∞∞ ,-15) and (4.40,19.0) f is increasing…
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- MATH
f(x)=x6−15×5+75×4−125×3−xf(x)=x6-15×5+75×4-125×3-x Produce graphs of ff that reveal all the…
f(x)=x6−15×5+75×4−125×3−xf(x)=x6-15×5+75×4-125×3-x f'(x)=6×5−75×4+300×3−375×2−1f′(x)=6×5-75×4+300×3-375×2-1 f”(x)=30×4−300×3+900×2−750xf′′(x)=30×4-300×3+900×2-750x 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 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 ⇔⇔ x=0.5 ,…
1 educator answer
- MATH
2sec2(x)+tan(x)−6=0,[−π2,π2]2sec2(x)+tan(x)-6=0,[-π2,π2] Use a graphing utility to approximate the…
See the attached graph. x ≈≈ -1.050 , x ≈≈ 0.875
1 educator answer
- MATH
See the attached graph, x ≈≈ 1.100
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- MATH
cos2(x)−2cos(x)−1=0cos2(x)-2cos(x)-1=0 using quadratic formula, cos(x)=−(−2)±(−2)2−4⋅1⋅(−1)−−−−−−−−−−−−−−−−−−√2cos(x)=-(-2)±(-2)2-4⋅1⋅(-1)2 cos(x)=2±8–√2cos(x)=2±82 cos(x)=1±2–√cos(x)=1±2 Solutions for cosx=1+2–√2 : None as cos(x) can not be…
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- MATH
see the attached graph Solutions of equation in the interval −π2≤x≤π2-π2≤x≤π2 are x ≈≈ 0.525 , -1.150
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- MATH
csc2(x)−5csc(x)=0csc2(x)-5csc(x)=0 Use inverse functions where needed to find all solutions of the…
csc2(x)−5csc(x)=0csc2(x)-5csc(x)=0 is equivalent to csc(x)⋅(csc(x)−5)=0,csc(x)⋅(csc(x)-5)=0,so csc(x)=0orcsc(x)=5.csc(x)=0orcsc(x)=5. By definition, csc(x)=1sin(x)csc(x)=1sin(x) and this is never zero. So sin(x)=15sin(x)=15 remains. There are two…
1 educator answer
- MATH
csc2(x)+3csc(x)−4=0csc2(x)+3csc(x)-4=0 Use inverse functions where needed to find all solutions of…
This is a quadratic equation for csc(x). The roots are 1 and -4. csc(x)=1/sin(x), so we have two possibilities: sin(x)=1 or sin(x)=-1/4. The roots of these equations in (0,2π)(0,2π) are π2,π2,…
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- MATH
sec2(x)+2sec(x)−8=0sec2(x)+2sec(x)-8=0 Use inverse functions where needed to find all solutions of…
Denote sec(x) as y and obtain a quadratic equation for y: y2+2y−8=0.y2+2y-8=0. This equation has two roots, y1=2y1=2 and y2=−4.y2=-4. So we have to solve sec(x)=2 and sec(x)=-4 separately. sec(x) = 1/cos(x),…
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- MATH
sec2(x)−4sec(x)=0sec2(x)-4sec(x)=0 Use inverse functions where needed to find all solutions of the…
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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- MATH
cot2(x)−6cot(x)+5=0cot2(x)-6cot(x)+5=0 Use inverse functions where needed to find all solutions of…
This is a quadratic equation for cot(x). There are two roots, cot(x) = 1 and cot(x)=5. Now solve these equations. On (0,2π)(0,2π) cot(x)=1cot(x)=1 at x1=π4×1=π4 and x2=5π4.×2=5π4. Also cot(x)=5 at…
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- MATH
cot2(x)−9=0cot2(x)-9=0 Use inverse functions where needed to find all solutions of the…
cot2(x)=9,cot2(x)=9, cot(x)=3 or cot(x)=-3. The general solution for cot(x)=a is On (0,2π)(0,2π) there are cot−1(3),cot−1(3)+π,π−cot−1(3),2π−cot−1(3).cot-1(3),cot-1(3)+π,π-cot-1(3),2π-cot-1(3). Note that cot−1(−3)=−cot−1(3).cot-1(-3)=-cot-1(3).
1 educator answer
- MATH
First transform cos2(x)cos2(x) into 1−sin2(x)1-sin2(x) : 2−2sin2(x)+7sin(x)=5,2-2sin2(x)+7sin(x)=5, 2sin2(x)−7sin(x)+3=0.2sin2(x)-7sin(x)+3=0. This is a quadratic equation for sin(x), the roots are sin(x)=3 and sin(x)=1/2. The first is…
1 educator answer
- MATH
First transform sin2(x)sin2(x) into 1−cos2(x)1-cos2(x) and obtain 2−2cos2(x)+5cos(x)=4,2-2cos2(x)+5cos(x)=4, 2cos2(x)−5cos(x)+2=0.2cos2(x)-5cos(x)+2=0.This is a quadratic equation for cos(x), the roots are 2 and 1/2. cos(x) never =2….
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- MATH
sec2(x)+tan(x)−3=0sec2(x)+tan(x)-3=0 Use inverse functions where needed to find all solutions of…
sec2(x)+tan(x)−3=0sec2(x)+tan(x)-3=0 use the identity sec2(x)=1+tan2(x)sec2(x)=1+tan2(x) 1+tan2(x)+tan(x)−3=01+tan2(x)+tan(x)-3=0 tan2(x)+tan(x)−2=0tan2(x)+tan(x)-2=0 solve using quadratic formula, tan(x)=−1±12−4⋅1⋅(−2)−−−−−−−−−−−−−−√2tan(x)=-1±12-4⋅1⋅(-2)2…
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- MATH
First, transform sec2(x)sec2(x) into 1+tanx1+tanx (this is obvious): 1+tan2(x)−6tan(x)=−4,1+tan2(x)-6tan(x)=-4, or tan2(x)−6tan(x)+5=0.tan2(x)-6tan(x)+5=0.This is a quadratic equation for tan(x), its roots are 1 and 5. So we have…
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- MATH
tan2(x)−tan(x)−2=0tan2(x)-tan(x)-2=0 Use inverse functions where needed to find all solutions of…
This is the quadratic equation for tan(x). It has roots -1 and 2. tan(x)=-1 on (0,2π)(0,2π) at x1=3π4×1=3π4 and x2=7π4.×2=7π4. tan(x)=2 on (0,2π)(0,2π) at x3=tan−1(2)x3=tan-1(2) and x4=π+tan−1(2).x4=π+tan-1(2)….
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- MATH
tan2(x)+tan(x)−12=0tan2(x)+tan(x)-12=0 Use inverse functions where needed to find all solutions of…
tan2(x)+tan(x)−12=0tan2(x)+tan(x)-12=0 using quadratic formula, tan(x)=−1±12−4⋅1⋅(−12)−−−−−−−−−−−−−−−√2tan(x)=-1±12-4⋅1⋅(-12)2 tan(x)=−1±49−−√2=−1±72=3,−4tan(x)=-1±492=-1±72=3,-4 solutions for tan(x)=3 for the range 0≤x≤2π0≤x≤2π
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- MATH
4cos2(x)−4cos(x)−1=04cos2(x)-4cos(x)-1=0 Use the Quadratic Formula to solve the equation in the…
4cos2(x)−4cos(x)−1=04cos2(x)-4cos(x)-1=0 using quadratic formula, cos(x)=−(−4)±(−4)2−4⋅4⋅(−1)−−−−−−−−−−−−−−−−−−√2⋅4cos(x)=-(-4)±(-4)2-4⋅4⋅(-1)2⋅4 cos(x)=4±16+16−−−−−−√8cos(x)=4±16+168 cos(x)=1±2–√2cos(x)=1±22 For cos(x)=1+2–√21+22 , No solution since…
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- MATH
tan2(x)+3tan(x)+1=0tan2(x)+3tan(x)+1=0 Use the Quadratic Formula to solve the equation in the…
tan2(x)+3tan(x)+1=0tan2(x)+3tan(x)+1=0 using quadratic equation formula, tan(x)=−3±32−4⋅1⋅1−−−−−−−−−−√2tan(x)=-3±32-4⋅1⋅12 tan(x)=−3±9−4−−−−−√2tan(x)=-3±9-42 tan(x)=−3±5–√2tan(x)=-3±52 Solutions for tan(x)=−3−5–√2tan(x)=-3-52 for the range…
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- MATH
3tan2(x)+4tan(x)−4=03tan2(x)+4tan(x)-4=0 Use the Quadratic Formula to solve the equation in the…
3tan2(x)+4tan(x)−4=03tan2(x)+4tan(x)-4=0 using quadratic formula, tan(x)=−4±42−4⋅3⋅(−4)−−−−−−−−−−−−−−√2⋅3tan(x)=-4±42-4⋅3⋅(-4)2⋅3 tan(x)=−4±16+48−−−−−−√6tan(x)=-4±16+486 tan(x)=−4±86=−2,23tan(x)=-4±86=-2,23solutions for tan(x)=-2 for the range…
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- MATH
12sin2(x)−13sin(x)+3=012sin2(x)-13sin(x)+3=0 Use the Quadratic Formula to solve the equation in the…
12sin2(x)−13sin(x)+3=012sin2(x)-13sin(x)+3=0 using quadratic formula, sin(x)=−(−13)±(−13)2−4⋅12⋅3−−−−−−−−−−−−−−−−√2⋅12sin(x)=-(-13)±(-13)2-4⋅12⋅32⋅12 sin(x)=13±169−144−−−−−−−−√24sin(x)=13±169-14424 sin(x)=13±25−−√24sin(x)=13±2524 sin(x)=13±524=34,13sin(x)=13±524=34,13 Solutions of…
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- MATH
see attached graph. Solutions in the given interval are, x ≈≈ 0.500 , 0.700 , 2.400 , 2.600
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- MATH
2tan2(x)+7tan(x)−15=02tan2(x)+7tan(x)-15=0 Use a graphing utility to approximate the solutions (to…
See attached graph Solutions in the given interval are, x ≈≈ 1.000 , 1.750 , 4.125 , 4.875
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- MATH
csc2(x)+0.5cot(x)−5=0csc2(x)+0.5cot(x)-5=0 Use a graphing utility to approximate the solutions (to…
See the attached graph. Solutions in the given interval are, x ≈≈ 0.500 , 2.700 , 3.650 , 5.850
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