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What values of c satisfy the Mean Value Theorem of y=-(6x 24)^(2
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y=-(6x+24)^(2/3); [-4, -1] | Find the values of c that satisfy the Mean Value Theorem
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y=-(-5x+25)^(1/2); [3, 5] | Find the values of c that satisfy the Mean Value Theorem
0:02:09
y=(x-3)^(2/3); [1, 4] | Find the values of c that satisfy the Mean Value Theorem
0:03:07
y=x^3+3x^2-2; [-2, 0] | Find the values of c that satisfy the Mean Value Theorem
0:01:57
y=-x^2/(4x+8); [-3, -1] | Find the values of c that satisfy the Mean Value Theorem
0:03:46
y=(-x^2+9)/4x; [1, 3] | Find the values of c that satisfy the Mean Value Theorem
0:02:24
y=-(x^2/2)+x-1/2; [-2, 1] | Find the values of c that satisfy the Mean Value Theorem
0:03:08
Find all numbers c that satisfies the Mean Value Theorem f = 6/x - 3 on [-1, 2]
0:03:32
y=(x^2-9)/3x; [1, 4] | Find the values of c that satisfy the Mean Value Theorem
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y=-x^3+4x^2-3; [0, 4] | Find the values of c that satisfy the Mean Value Theorem
0:02:10
y=(x^2/2)-2x-1; [-1, 1] | Find the values of c that satisfy the Mean Value Theorem
0:03:12
Find all numbers c that satisfies the Mean Value Theorem f = x^3 -3x +2 on the interval [-2, 2]
0:02:44
Find all numbers c that satisfies the Mean Value Theorem f = 3x^2 + 5x -2 on the interval [-1, 1]
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Mean Value Theorem for Integrals | Ex. 4 of 16 | f(x)=4/x^2; [-4,-2]
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How REAL Men Integrate Functions
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Find all numbers c that satisfies the Mean Value Theorem f = x^3 + 12x^2 + 7x on [-4, 4]
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Human Calculator Solves World’s Longest Math Problem #shorts
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Find the open intervals on which the function 23. ƒ(Θ) = 3Θ^2 - 4Θ^3 24. ƒ(Θ) = 6Θ - Θ^3
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IIT Bombay CSE 😍 #shorts #iit #iitbombay
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Find the value or values of c 7. ƒ(x) = x^3-x^2 ,[-1,2] 8.g(x)=x^3,- 2≤x≤0;x^2,0<x≤2
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Find the open intervals on which the function 21. h(x) = -x^3 + 2x^2 22. h(x) = 2x^3 - 18x
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Identify the function’s local extreme values49. ƒ(t) = 12t - t^3,-3≤t<inf;50.ƒ(t)=t^3-3t^2,-inf<t≤3
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Modulo Operator Examples #Shorts #math #maths #mathematics #computerscience
0:02:37
Which of the functions in Exercises 9–14 satisfy the 9. ƒ(x) = x^(2/3),[-1, 8]10.ƒ(x)=x^(4/5), [0,1]
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