# Math Homework Help

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• Math
Hello! Consider this function for t in [-1, 3] and [3, 5] separately. At the first interval t<=3 and |t-3| = -(t-3) = 3-t.So y(t) = 3 - (3-t) = t.The maximum is at the maximum t, i.e. at t=3,...

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• Math
Hello! For x<=-4 x+4<=0 and g(x) = -(x+4).The minimum on the interval [-7, -4] is at the x=-4, g(-4) = 0.The maximum on the interval [-7, -4] is at the x=-7, g(-7) = 3. For x>=-4...

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• Math
f(x) = [x], is the nearest integer function. For values of x in the closed set [-2, 2], the maximum value of f(x) is 2 and the minimum value of f(x) is -2. The absolute extrema of the function are...

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• Math
The function h(x) = [2-x] is the nearest integer function. For values of x lying in the closed set [-2,2], the maximum value of h(x) = 2 - (-2) = 2 + 2 = 4. The minimum value of x is 2 - 2 = 0 The...

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• Math
Given the function: `f(x)=sin(x)` in the closed interval [5 pi/6 , 11 pi/6] We have to find the absolute extrema of the function on the closed interval. Step 1: Check whether f(x) is a continuous...

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• Math
Given the function g(x)=sec(x) in the interval [-pi/6, pi/3] We have to find the absolute extrema of the function on the closed interval. So first let us find the derivative of the function and...

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• Math
Given: `y=3cos(x),[0,2pi]` First find the critical x values of the function. To find the critical values, set the derivative equal to zero and solve for the x value(s). `f'(x)=-3sin(x)=0`...

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• Math
Given the function `y=tan((pi x)/8)` in [0,2]. We have to find the absolute extrema of the function in the closed interval [0,2]. So first we have to take the derivative of the function and equate...

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• Math
You need to evaluate the critical numbers of the function and for this reason, you must differentiate the function with respect to x, such that: `f'(x) = (x^3 - 3x^2)'` `f'(x) = 3x^2 - 6x` You need...

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• Math
You need to evaluate the critical numbers of the function and for this reason, you must differentiate the function with respect to x, such that: `f'(x) = (x^4 - 8x^2)'` `f'(x) = 4x^3 - 16x` You...

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• Math
You need to evaluate the critical numbers of the function and for this reason, you must differentiate the function with respect to t, using the product and chain rules, such that: `g'(t) =...

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• Math
Given `f(x)=(4x)/(x^2+1)` Find the derivative using the Quotient Rule. The set the derivative equal to zero and solve for the critical x value(s). `f'(x)=[(x^2+1)(4)-(4x)(2x)]/(x^2+1)=0`...

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• Math
Given the function `h(x)=sin^2(x)+cos(x)` in the interval `0<x<2pi` We have to find the critical numbers of the function. First take the derivative of the function and equate it to zero. We...

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• Math
Given the function f(x) =2sec(`theta)`+tan(`theta)` in the interval 0<theta<2 pi We have to find the critical points. So let us first take the derivative of the function and evaluate it to...

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• Math
Given: f(x)=3-x,[-1,2] First find the critical values of the function. To find the critical values of the function (if one exists), set the derivative of the function equal to zero and solve for...

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• Math
Given: `f(x)=(3/4)x+2,[0, 4]` First find the critical value(s) of the function (if any exist). To find the critical value(s) of the function, set the derivative equal to zero and solve for the...

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• Math
Given: `g(x)=2x^2-8x,[0,6]` First find the critical x value(s) of the function. You will do this by setting the derivative function equal to zero and solving for the x value(s). `g'(x)=4x-8=0`...

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• Math
Given: `h(x)=5-x^2, [-3, 1]` Find the critical values of the function by setting the derivative equal to zero and solving for x. When the derivative is equal to zero the slope of the tangent line...

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• Math
Given: `f(x)=x^3-(3/2)x^2,[-1,2]` First find the critical x value(s) of the function. To find the critical x value(s), set the derivative equal to zero and solve for the x value(s)....

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• Math
`f(x) = 4/(x^2 + 1)` `or, f(x) = 4*(x^2 + 1)^-1` `thus, f'(x) = -4*(2x)*(x^2 +1)^-2` `or, f'(x) = -8x/(x^2 +1)^2` `or, f'(-1) = (-8*-1)/{(-1)^2 + 1}^2` `or, f'(x) = 8/4 = 2` ``

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• Math
`f(x) = (3x+1)/(4x-3)` `f'(x) = [3*(4x-3) - 4*(3x+1)]/(4x-3)^2` `or, f'(x) = -13/(4x-3)^2` `or, f'(4) = -13/(4*4 - 3)^2` `or, f'(4) = -13/(13)^2` `or, f'(4) = -1/13` ``

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• Math
`y = (1/2)*cosec(2x)` `y' = -(1/2)*2*cosec(2x)*cot(2x)` Putting x = pi/4 `y' = -1*cosec(pi/2)*cot(pi/2) = -1*1*0 = 0` ``

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• Math
`y = cosec(3x) + cot(3x)` `y' = -3*cosec(3x)*cot(3x) - 3*cosec^2(3x)` Putting x = pi/6 we get y' = `-3*cosex(pi/2)*cot(pi/2) - 3*cosec^2(pi/2)` `or, y' = -3*1*0 - 3*1` `or, y' = -3` ``

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• Math
`y = (8x+5)^3` `y' = 3*8*(8x+5)^2` `y' = 24*(8x+5)^2` `y'' = 24*2*8*(8x+5)` `or, y'' = 384*(8x+5)` `` ` `

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• Math
`y = 1/(5x+1) = (5x+1)^-1` ` ` `differentiating` `y' = -1{(5x+1)^-2}*5` `or, y' = -5(5x+1)^-2` ``Differentiating againg w.r.t 'x' we get `y'' = 10*5*(5x+1)^-3` `or, y'' = 50/(5x+1)^3` ``

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• Math
Note:- 1) If y = cotx ; then dy/dx = `-cosec^2(x)` ` ` 2) If y = cosecx ; then dy/dx = -cosecx*cotx Now, `f(x) = y = cotx` `differentiating ` `f'(x) = y' = -cosec^2(x)` `differentiating` `f''(x) =...

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• Math
Note:- If y = sinx ; then dy/dx = cosx If y = cosx ; then dy/dx = -sinx 2sinx*cosx = sin(2x) Now, `y = (sinx)^2` `or, y = sin^2x` `differentiating` `y' = 2sinx*cosx` `or,y' = sin(2x)`...

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• Math
`(x^2) + (y^2) = 64` `Differentiating` `2x + 2y(dy/dx) = 0` `or, x + y(dy/dx) = 0` `or, dy/dx = -x/y` ``

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• Math
`(x^2) + 4xy - (y^3) = 6` `differentiating` `2x + 4y + 4x(dy/dx) - 3(y^2)*(dy/dx) = 0` `or, 2(x+2y) = (dy/dx)*[3(y^2) - 4x]` `or, dy/dx = [2(x+2y)]/[3(y^2)-4x]` ``

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• Math
`(x^3)*y - x*(y^3) = 4` `differentiating ` `3(x^2)*y + (x^3)*(dy/dx) - (y^3) - 3x(y^2)*(dy/dx) = 0` `or, (dy/dx)*[(x^3)-3x(y^2)] = (y^3) - 3(x^2)y` `or, dy/dx = [(y^3) - 3(x^2)y]/[(x^3)-3x(y^2)]` ``

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• Math
Note:- If y = x^n ; where 'n' = constant ; then dy/dx = n*x^(n-1) Now, `(x*y)^(1/2) = x - 4y` `or, {x^(1/2)}*{y^(1/2)} = x - 4y` `or, (1/2)*{x^(-1/2)}*{y^(1/2)} +...

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• Math
Note:- 1) If y = sinx; then dy/dx = cosx 2) If y = cos(x) ; then dy/dx = -sinx Now, `x*sin(y) = y*cos(x)` `or, x*cosy*(dy/dx) + sin(y) = -y*sin(x) + cos(x)*(dy/dx)` `or, x*cos(y)*(dy/dx) -...

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• Math
Note:- If y = cos(x) ; then dy/dx = -sin(x) Now, `cos(x+y) = x` `or, -sin(x+y)*{1 + (dy/dx)} = 1` `or, 1+(dy/dx) = -1/sin(x+y)` `or, 1+(dy/dx) = -cosec(x+y)` `or, dy/dx = -1 - cosec(x+y)` `or,...

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• Math
Note:- If y = x^n ; where n = constant ; then dy/dx = n*{x^(n-1)} Now, `y = 15*x^(5/2)` `y' = 15*(5/2)*x^{(5/2)-1}` `or, y' = (75/2)*x^(3/2)` `thus, y'' = (75/2)*(3/2)*x^{(3/2)-1}` `or, y'' =...

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• Math
Note:- If y = x^n ; where n = constant ; then dy/dx = n*{x^(n-1)} Now, `y = 20*x^(1/5)` `y' = 20*(1/5)*x^{(1/5)-1}` `or, y' = 4*x^(-4/5)` `thus, y'' = 4*(-4/5)*x^{(-4/5)-1}` `or, y'' =...

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• Math
Note:- If y = tanx ; then dy/dx = sec^2(x) If y = sec(x) ; then dy/dx = sec(x)*tan(x) Now, `f(theta) = 3tan(theta)` `f'(theta) = 3sec^2(theta)` `f''(theta) = 3*2sec(theta)*sec(theta)*tan(theta)`...

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• Math
Note:- If y = cos(ax) ; then dy/dx = -a*sin(ax) If y = sin(ax) ; then dy/dx = a*cos(ax) ; where 'a' = constant Now, `h(t) = 10cos(t) - 15sin(t)` `h'(t) = -10sin(t) - 15cos(t)` `h''(t) = -10cos(t)...

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• Math
Note:- If y = x^n ; where n = constant ; then dy/dx = n*{x^(n-1)} Now, y = `(7x +3)^4` `thus, dy/dx = y' = 4{(7x+3)^3}*7` `or, dy/dx = y' = 28(7x+3)^3` ``

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• Math
Note:- If y = x^n ; where n = constant ; then dy/dx = n*{x^(n-1)} Now, `y = (x^2 - 6)^3` `thus, dy/dx = y' = 3*{(x^2 - 6)^2}*(2x)` `or, dy/dx = y' = 6x*(x^2 - 6)^2` ``

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• Math
Note:- If y = x^n ; where n = constant, then dy/dx = n*x^(n-1) Now, `y = 1/{(x^2) + 4}` `Thus, y = {(x^2) + 4}^-1` `or, y' = -1{{(x^2) + 4}^-2}*(2x)` `or, y' = -2x/{(x^2)+4}^2` ``

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• Math
Note :- 1) If y = x^n ; where n = constant, then dy/dx = n*(x^(n-1)) 2) If y = n*x ; where n = constant ; then dy/dx = n Now, `y = 1/(5x+1)^2` `or, y = (5x+1)^-2` `thus, dy/dx = y' =...

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• Math
Note:- If y = cos(ax) ; then dy/dx = -a*sin(ax) Now, `y = 5cos(9x + 1)` `dy/dx = y' = -5*(sin(9x+1))*9` `or, dy/dx = y' = -45sin(9x+1)` ``

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• Math
y = 1- cos(2x) + 2`(cos^2x)` `dy/dx = y' = 2*sin(2x) - 4*cosx*sinx` `or, y' = 2sin(2x) - 2sin(2x) = 0` note: `2sinx*cosx = sin2x`

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• Math
Note:- 1) If y = sin(x) ; then dy/dx = cos(x) 2) If y = x^n ; then dy/dx = n*x^(n-1) ; where n = constant Now, `y = (x/2) - {sin(2x)/4}` `dy/dx = y' = (1/2) - {2cos(2x)}/4` `or, dy/dx = y' = (1/2)...

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• Math
Given: `y=(sec(x))^7/7-(sec(x))^5/5` `y'=(7sec(x)^6)/7sec(x)tan(x)-(5sec(x)^4)/5sec(x)tan(x)` `y'=(sec(x))^7tan(x)-(sec(x))^5tan(x)` `y'=(sec(x))^5tan(x)[(sec(x))^2-1]`...

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• Math
You need to differentiate the function with respect to x, using the product rule and chain rule, such that: `y' = x'*(6x + 1)^5 + x*((6x + 1)^5)'` `y' = 1*(6x + 1)^5 + x*5*(6x + 1)^4*(6x+1)'` `y' =...

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• Math
`f(s)=(s^2-1)^(5/2)(s^3+5)` Using the product rule for the derivative, `f'(s)=(s^2-1)^(5/2)d/(ds)(s^3+5)+(s^3+5)d/(ds)(s^2-1)^(5/2)` `f'(s)=(s^2-1)^(5/2)(3s^2)+(s^3+5)(5/2)(s^2-1)^((5/2)-1)(2s)`...

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• Math
`f(x)=3x/sqrt(x^2+1)` Using the quotient rule of the derivative, `f'(x)=3((sqrt(x^2+1)-xd/(dx)(sqrt(x^2+1)))/((sqrt(x^2+1))^2))` `f'(x)=3((sqrt(x^2+1)-x(1/2)(x^2+1)^(-1/2)2x)/(x^2+1))`...

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• Math
Given: `h(x)=((x+5)/(x^2+3))^2` To find the derivative of the function use the Quotient Rule within the Chain Rule. `h(x)=2((x+5)/(x^2+3))[((x^2+3)(1)-(x+5)(2x))/(x^+3)^2]`...

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