# Solution Essay Examples

## Problem Solving

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Solution. tan-1x = arctan x at the interval (-π,2π,2) therefore the horizontal asymptote for 2 arctan x is y = ±π for y=2tan-1x(e) Asymptotes of f(x) =9x2+59x4+1dropping dominant terms we drop everything except big exponent f(x) =9x29x4=9x23x2=3Hence the horizontal asymptote equation is y=33). Consider f(x)=x2-2x+4x-2a). the domain. x<2b). the coordinates of the x-intercepts if y=0, x is not defined hence no solutions c). coordinates of the y-intercepts 0,-2d).the coordinates of the critical points (0,-2) and (4, 6) e). the intervals on which the function increases or decreases get the first derivative using the quotient rule (Chen, Fox & Lyndon 82)...

• Words: 825
• Pages: 3

## HOME WORK

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Solution Given dl/dt=3 And dw/dt=-2 Then dp when l=12 ft and w=8ft Perimeter= 2l+2w Dp/dt=d/dt{2l+2w} =2dl/dt+2dw/dt=2(3)+2(-2) =2 ft/sec How fast is the area changing? Solution Given dl/dt=3, dw/dt=-2 Da/dt when l=12 ft , w=8ft Area=lxwDa/dt=d/dt{lxw} =Dl/dt*w+dw/dt*l =3(8)+ 12(-2) =0 ft2/sec (c) Find all critical points and local extremes of the following function on the given intervals. f (x) = 2-x3 on Solution To find the critical points, take derivative: f’(x)=-3x2 f’(X)= 0 0=-3x2 X=0 For absolute minima and maxima; f(0)=2-(0)3=2 f(-2)=2-(-2)3=10 f(1)=2-(1)3=1 Therefore: abs minima =1 at x=1 Abs maxima=10 at x=-2 (d) Calculate the limits of the...

• Words: 275
• Pages: 1

## 983176648 test 3 revised

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solution of by using the following steps (i) Find, where xn+1=xn- f(x)f'(x)f'x= 6x²+5x0 f(x)f'(x)xn+1=xn- f(x)f'(x)0-651.2x=1.2ii) Find, where xn+1=xn- f(x)f'(x)f'x= 6x²+5x0 f(x)f'(x)x0+1=x0- f(x)f'(x)11110.0909x=0.0909iii)x0 = 3 iv)Find, where xn+1=xn- f(x)f'(x)f'x= 6x²+5x0 f(x)f'(x)x0+1=x0-...

• Words: 550
• Pages: 2

## Algebra 2 Radicals and Complex Numbers

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Solution: 4x23y is the same as 4*√x23y= 2x3ySimplify the expression. Show your work. Solution: Opening the brackets; 3*1+3*5i-4i*1 -4i*5i-2-iCollecting like terms; 3+15i-4i-20i2-2-i3+11i-20i2-2+i 1+12i-20i2The formula represents the length of the hypotenuse of a right triangle with side lengths a and b. Solve the equation for b. Show your work. Solution: Finding the square of both sides gives: c2= a2+b2b2= c2- a2Now find the square roots in both sides. Therefore, b...

• Words: 275
• Pages: 1