Wednesday

August 27, 2014

August 27, 2014

Posted by **Samantha** on Thursday, December 19, 2013 at 11:58pm.

1. If g(x) = tan (5x^2), then g′(x) =

* 10x sec2(5x2)

sec2(5x2)

–10xsec(5x2)tan(5x2)

10sec2(5x2)

7xsec2(5x2)

2. The inflection point of the curve y = x4 − 8x3 + 24x2 + 7x − 3 is:

(−1, 23)

(0, 3)

(1, 21)

*(2, 59)

There is no point of inflection.

3. Suppose the functions f and g and their derivatives have the following values at x = 1 and x = 2. Let h(x) = f(g(x)). Evaluate h′(1).

x|f(x) g(x) f'(x) g'(x)

--------------------------

1| 8 2 1/3 -3

2| 3 -4 2pi 5

-6π

* -15

-1

10

24

4. The slope of the tangent line to the graph y=(x^3/3)-x at the point (1,-2/3)is:

-1

0

2

* -2/3

undefined

5. The derivative of y=cosx/1+sinx is:

-sinx/cosx

tan x

cos^2x-sin^2x/(1+sinx)^2

-1/(1+sinx)

* sinx/-cosx

6. Which of the following functions is continuous but not differentiable at x = 1?

I. y=^3(sqrt x-1)

II. y={x^2, x ≤ 1

{2x, x > 1

III. y={x, x ≤ 1

{1/2, x > 1

I only

II only

* I and III only

II and III only

All of these functions are continuous but not differentiable at x = 1.

- Check a few more CALC questions, please? -
**Steve**, Friday, December 20, 2013 at 12:12am#1. Since d/dx tan x = sec^2 x, you have

10x sec^2(5x^2)

#2. ok

#3. ok

#4. y' = x^2-1

so, 0

#5.

y = cos(1+sin)

y' = [(-sin)(1+sin)-cos(cos)]/(1+sin)^2

= (-sin-sin^2-cos^2)/(1+sin)^2

= (-sin-1)/(1+sin)^2

= -1/(1+sin)

#6.

I. No idea what ^3(sqrt x-1) means

II. y(1) = 1 or 2, so not continuous

III. y(1) = 1 or 1/2, so not continuous

So, only possible choice is I or none.

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