Consider the function f(x) = - +4. In this problem you will calculate √(- ³²3² + +4) dx by using the definition Sº f a f(x) dx = n lim f(x;) Ax n→∞ i=1

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Q: nd the points where the function f(x) = (x² – 1)³ is decreasing, -

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Q: - h)? + k: a). f(x) = 2x² –12x+11; b) y = -r+x+:

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Q: Suppose that f(x + h) – f(x) -2h(x + 3) – h? h(x + h + 3)²(x + 3)² Find the slope m of the tangent…

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Q: Given the function f(x) = x² – 3x f(2+h)-f(2) h (a) Find the expression (b) Use your result from…

A: The given function is ; f(x)=x2-3x.

Q: Let f be the function given by f(x)=√x² - 9x² Find f'(x).

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Q: By carefully showing all your steps, differentiate the function f(x) = (tan−¹(x + √√x)) 4

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Q: f(x)=√x −7 lim h→0 f(6+h)−f(6)/h

A: Question is solved.

Q: Find Lim-0.0 (r² + y²) In(x² +y²) '(xy)(0,0)

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Q: Find f"(x) for the function: f(x) = x – 1. -2(x– 1)-2 2(x – 1)2 O 2(x – 1)-3 -2(x– 1)

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Q: Find f(x) if f (x) = x/x given f(1) = 2. %3D

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Q: Ex: f(x) = 3x+8 and g(x)=2x-1 Find (f/g)(x). 3x+8 2x-1' 13 24 2x-1 3x+8' 2x-1 3x+8' 80 x # 1/1/2 x =…

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Q: 'Зx2 + 8х + 4) lim п--2 х2 + 3x + 2

A: Given query is to find the value of expression at indicated limit.

Q: 2x + Vx. 4. Determine a function f (x) knowing that f"(x) = 16e¬

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Q: An inflection point of the function ƒ (x) = 2x³ – 12x² – 24x −— 10 is at x= 01/1/20 O 01 O 3 04

A: The inflection point is a point in which the concavity of the function changes. It means that the…

Q: (A) (B) ƒ(5) - ƒ(3) 5-3 f(3 +h)-f(3) h (C) lim h→0 || f(3+h)-f(3) h

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Q: Consider the function g(x)=(f(x))^5. If f(4)=3 and f′(4)=−2, then find g′(4).

A: Given: g(x)=f(x)5f(4)=3f'(4)=-2

Q: Let f(x) = 3+ √5-8. Find f f(₂)=[

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Q: Let f(x)=3−x2. Find each of the following: a) f(−1)−f(−6)−1−(−6)=  b) f(−6+h)−f(−6)h=  c)…

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Q: Can f(z) = | Re (z2) | + i.Im (-z) be derived from the function z=0? Show it.

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Q: What input is NOT allowed for f(x)= 1/5x-25

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Q: If f3 g(x) dx = 12 and ₁ g(x) dx =-5, then f¹3 -3 g(x) dx = A)-7 B)-17 C) 7 D) 17

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Q: For the function f(x) = 4x? +16x + 25, find f'(x) using the definition f(x+ h) - f(x) lim h-0

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