**hey guys its charlene. i would of posted this earlier but then it got lost becuase i didn't save it lol. i cant put pictures of anything up because it won't let me. so please just bare with me..**

alright so today we had our pre test on chapter 2 : derivative functions

1) if f(x) ln square root X, then the average rate of change of f on the intercal [3,7] is approxiamately:

..............a) 0.106 ........b) 0.189 ......c) 0.212...... d) 0.424 .....e) 0.847

- you graph the function on your calculator

- find the values for X=3 and X=7

-plot these values into the slope formula

-when you do this you get " 0.106 "

alright so today we had our pre test on chapter 2 : derivative functions

1) if f(x) ln square root X, then the average rate of change of f on the intercal [3,7] is approxiamately:

..............a) 0.106 ........b) 0.189 ......c) 0.212...... d) 0.424 .....e) 0.847

- you graph the function on your calculator

- find the values for X=3 and X=7

-plot these values into the slope formula

-when you do this you get " 0.106 "

**2) suppose the number of bacteria in a certain organism grows over time and the number, N(t), of bacteria (measured in thousands) at a time t is given by**

N(t) = t(2+cost)^(4/3) + 3t

at approximately what time, t, in the first 10 days is the number of bacteria growing the fastest?

.............a) 5.0 .......b)5.6 .........c)5.12......... d) 5.18 .......e)5.24

-you can graph the function into you calculator

- you can use NDERIV to find the slopes at each value eg. 5.0 , 5.6

-the point with the largest slope is the answer, in this case it is 5.18

N(t) = t(2+cost)^(4/3) + 3t

at approximately what time, t, in the first 10 days is the number of bacteria growing the fastest?

.............a) 5.0 .......b)5.6 .........c)5.12......... d) 5.18 .......e)5.24

-you can graph the function into you calculator

- you can use NDERIV to find the slopes at each value eg. 5.0 , 5.6

-the point with the largest slope is the answer, in this case it is 5.18

**3) lim**

__2x__^{2+X-3}

X->1 3x

a) 2/3 b)3/2 c)1 d) 5.18 e)5.24

-factor out the function

lim

X->1 (3X+2)(X-1)

-the (X-1) cancel out

-plug 1 into the remaining function and then you get an answer of 1

X->1 3x

^{2}-X-2a) 2/3 b)3/2 c)1 d) 5.18 e)5.24

-factor out the function

lim

__(2X+3)(X-1)__X->1 (3X+2)(X-1)

-the (X-1) cancel out

-plug 1 into the remaining function and then you get an answer of 1

**4) the graph of the second derivative of a function f is shown at right. which of the following is true?**

I) the graph of f has an inflection point at X=-1

II) the f-graph is concave down on the interval (-1,3)

III) the graph of the derivative function f ' is increasing at X=1

a) 2/3 b)II only c) III only d) I and II only e) I, II, III

I) the graph of f has an inflection point at X=-1

II) the f-graph is concave down on the interval (-1,3)

III) the graph of the derivative function f ' is increasing at X=1

a) 2/3 b)II only c) III only d) I and II only e) I, II, III

**5) suppose a function f is defined so that it has derivatives:**

f ' (x) = x

over what interval is the graph of f both increasing and conave up?

.................a) x less than o b) o<2/3.......c)2/3< x <1<>.......d) x>1......e) none of these

f ' (x) = x

^{2}(x-1) and f ' (x) = x(3x - 2)over what interval is the graph of f both increasing and conave up?

.................a) x less than o b) o<2/3.......c)2/3< x <1<>.......d) x>1......e) none of these

**6) consider the following table of data for the function f**

ill post the table later because it wouldn't let me sry

a) estimate f ' (5.2) as accurately as possible

you can use the slope formula and calculate the slope one unit before and after. then you find the average of those two slopes

.............(8.3-8.8) / (5.4 -5.2) = -2.5.......... (8.8-9.2) / (5.2-5.0) = -2

ill post the table later because it wouldn't let me sry

a) estimate f ' (5.2) as accurately as possible

you can use the slope formula and calculate the slope one unit before and after. then you find the average of those two slopes

.............(8.3-8.8) / (5.4 -5.2) = -2.5.......... (8.8-9.2) / (5.2-5.0) = -2

**..................................................**

**.....................................................(2.5) + (-2) / 2**

**...........................................................=2.25**

b) write the equation for the tangent line at x=5.2

- use the point slope formula

.............y-y1 = m (x-x1)

b) write the equation for the tangent line at x=5.2

- use the point slope formula

.............y-y1 = m (x-x1)

^{.............y-8.8 = -2.25 (x-5.2)}^{...................y = -2.25 (x-5.2) + 8.8}^{}^{c) use your answer to part (b) to find approximately, f(5.26)}^{-just plug 5.26 into the formula}**............y(5.26) -8.8 = 2.25 (5.26-5.2)**

**....................y(5.26) = -2.25(0.06) + 8.8**

**....................y(5.26) = 8.665**

**d) what is the sign of f '' (5.2)? explain your answer**

**the sign of f '' is NEGATIVE**

**-the slope of 5.0 to 5.2 is -2.0**

**-the slope of 5.2 to 5.4 is -2.5**

**-therefore the secant lines are decreasin**

**- therefore f ' is decreasing and f '' is less than zero on this this interval **

**THE NEXT SCRIBE IS JAN**

## No comments:

Post a Comment