AP Calculus Review #3

If the function f is continuous at x = 3 and $\lim_{x\rightarrow3^-}f'\left(x\right)=\lim_{x\rightarrow3^+}f'\left(x\right)$  , then which of the following must be true?

I.   $\lim_{x\rightarrow3}f\left(x\right)=3$  
II.  f is differentiable at x = 3

• II only 

• I only

• Both I and II

• Neither I or II

Answer explanation

The function is continuous as stated and the left and right hand limits of the derivative at x = 3 are equal, so it must also be differentiable at x =3.

The region R is the area enclosed by the functions $y=2x$  and  $y=x^2$  as shown.  Find the volume of the solid when the region R is rotated about the horizontal line y = -1.

•  $\pi\int_0^2\left(\left(x^2+1\right)^2-\left(2x+1\right)^2\right)dx$  

•  $\pi\int_0^2\left(\left(2x+1\right)^2-\left(x^2+1\right)^2\right)dx$  

•  $\pi\int_0^2\left(\left(x^2-1\right)^2-\left(2x-1\right)^2\right)dx$  

•  $\pi\int_0^2\left(\left(2x-1\right)^2-\left(x^2-1\right)^2\right)dx$  

Answer explanation

Top minus Bottom

A trapezoidal sum is an underestimate when the function is ...

• increasing

• concave down

• decreasing

• concave up

 $\frac{d}{dx}\left(e^{\frac{x}{4}}\right)=$  

•  $e^{\frac{x}{4}}$  

•  $-\frac{1}{4}e^{\frac{x}{4}}$  

•  $\frac{1}{4}e^{\frac{x}{4}}$  

•  $4e^{\frac{x}{4}}$  

Answer explanation

Chain rule!  Take the derivative of the exponent and throw it out front, then keep the e with the original exponent.

Tags

CCSS.HSF-IF.C.8B

Let  $g\left(x\right)=\int_0^xf\left(t\right)dt$  , where the graph of f is shown.  On what interval(s) is g both concave up and increasing?

• ( -2, 0 ) U ( 8, 10 )

• cannot be determined

• ( -2, 0 )

• ( -2, 0 ) U ( 5, 10)

Answer explanation

Since g'=f, g is concave up and increasing when f is increasing and positive.

For a particle moving along the x-axis,  $v\left(1\right)=-5$  and  $a\left(1\right)=2$  .  At time t = 1, it can be said that the particle is...

• moving away from the origin

• slowing down

• moving toward the origin

• speeding up

Answer explanation

Since  $\left|v\left(1\right)\right|>a\left(1\right)$  , the object must be slowing down.  The negative velocity just means the particle is moving in the negative direction.

For a particle moving along the x-axis,  $x\left(1\right)=5$  and  $v\left(1\right)=-10$  .  At time t = 1, it can be said that the particle is ...

• moving away from the origin

• moving toward the origin

• slowing down

• speeding up

Answer explanation

A particle moves toward the origin when the position and velocity are opposite signs.

 $\frac{d}{dx}\left(\frac{1}{2x+3}\right)=$  

•  $-\frac{2}{\left(2x+3\right)^2}$  

•  $2\ln\left|2x+3\right|$  

•  $\frac{1}{2}\ln\left|2x+3\right|$  

•  $\frac{2}{\left(2x+3\right)^2}$  

Answer explanation

Power rule and chain rule!

 $_{\int\cos x}$  =?

•  $-\sin x+c$  

•  $\frac{1}{\sqrt{1-x^2}}+c$  

•  $-\frac{1}{\sqrt{1-x^2}}+c$  

•  $\sin x+c$  

Answer explanation

Recall that the integral is an anti-derivative

If  $f\left(x\right)=\int_4^{x^2}\sin\left(2t\right)dt$  , then  $f'\left(x\right)=?$  

•  $4x\sin\left(2x^2\right)$  

•  $-2x\cos\left(2x^2\right)$  

•  $2x\sin\left(2x^2\right)$  

•  $-4x\cos\left(2x_{ }^2\right)$  

Answer explanation

When taking the derivative of an anti-derivative, take the derivative of the upper bound and throw it out front.  Then plug in the upper bound for the variable in the function.