If w ( x ) = f ( x ) q ( x ) w\left(x\right)=f\left(x\right)q\left(x\right) w ( x ) = f ( x ) q ( x ) , then w ′ ( x ) = w'\left(x\right)= w ′ ( x ) =

f ′ ( x ) q ( x ) + f ( x ) q ′ ( x ) f'\left(x\right)q\left(x\right)+f\left(x\right)q'\left(x\right) f ′ ( x ) q ( x ) + f ( x ) q ′ ( x )

f ′ ( x ) q ′ ( x ) f'\left(x\right)q'\left(x\right) f ′ ( x ) q ′ ( x )

f ′ ( x ) q ′ ( x ) + f ( x ) q ( x ) f'\left(x\right)q'\left(x\right)+f\left(x\right)q\left(x\right) f ′ ( x ) q ′ ( x ) + f ( x ) q ( x )

f ′ ( q ( x ) ) q ′ ( x ) f'\left(q\left(x\right)\right)q'\left(x\right) f ′ ( q ( x ) ) q ′ ( x )

Evaluate lim ⁡ x → π 4 cos ⁡ x − cos ⁡ ( π 4 ) x − π 4 \lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos x-\cos\left(\frac{\pi}{4}\right)}{x-\frac{\pi}{4}} x → 4 π lim x − 4 π cos x − cos ( 4 π )

− 2 2 -\frac{\sqrt{2}}{2} − 2 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 2 \frac{\sqrt{2}}{2} 2 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

For a particle moving along the x-axis, the particle's speed equals 2 when

∣ v ( t ) ∣ = 2 \left|v\left(t\right)\right|=2 ∣ v ( t ) ∣ = 2

v ( t ) = 2 v\left(t\right)=2 v ( t ) = 2

a ( t ) = 2 a\left(t\right)=2 a ( t ) = 2

∣ a ( t ) ∣ = 2 \left|a\left(t\right)\right|=2 ∣ a ( t ) ∣ = 2

If f ( x ) = ∫ 4 x ( cos ⁡ ( 2 t ) + 3 ) d t f\left(x\right)=\int_4^x\left(\cos\left(2t\right)+3\right)dt f ( x ) = ∫ 4 x ( cos ( 2 t ) + 3 ) d t , then f ′ ( x ) = f'\left(x\right)= f ′ ( x ) =

cos ⁡ ( 2 x ) + 3 \cos\left(2x\right)+3 cos ( 2 x ) + 3

1 2 sin ⁡ ( 2 x ) + 3 x \frac{1}{2}\sin\left(2x\right)+3x 2 1 sin ( 2 x ) + 3 x

cos ⁡ ( 2 x ) \cos\left(2x\right) cos ( 2 x )

2 cos ⁡ ( 2 x ) + 3 2\cos\left(2x\right)+3 2 cos ( 2 x ) + 3

What are the conditions that satisfy the mean value theorem, and what does it mean?

Continuous on the closed and differentiable on the open,and then there is at least one number c in the interval ( a , b ) (that is a < c < b ) such that f'(c)=(f(b)- f(a))/ (b-a)

Continuous on the open and differentiable on the closed, and then there is at least 2 numbers c and d in the interval ( a , b ) (that is a < c < b ) such that

Discontinuous on the closed and differentiable on the open, then there is at least one number c in the interval ( a , b ) (that is a < c < b ) such that f'(c)=(f(b)- f(a))/ (b-a)

Continuous on the open and differentiable on the open, and then there is no number c in the interval ( a , b ) (that is a < c < b ) such that f'(c)=(f(b)- f(a))/ (b-a)

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HSF.IF.A.2, 8.F.B.4, HSF.LE.B.5

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Tanya Hickson

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36 questions

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MULTIPLE CHOICE QUESTION

1 min • 1 pt

What is the limit?

Infinity

DNE

Find the second derivative of the function:
f (x) = 2x - 5x⁶

f ''(x)= 2 - 30x

f ''(x) = 2-30x⁵

f ''(x) = -30x⁵

f ''(x) = -150x⁴

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Intermediate Value Theorem

Rolle's Theorem

Average Rate of Change

Average Value of f

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Average Rate of Change

Average Value of f

Intermediate Value Theorem

Rolle's Theorem

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AP Calculus Final First 15 :)

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What is the limit?

Find the second derivative of the function:f (x) = 2x - 5x6

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Find the second derivative of the function:
f (x) = 2x - 5x⁶