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
3) lim 2x2+X-3
X->1 3x2 -X-2
a) 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
5) suppose a function f is defined so that it has derivatives:
f ' (x) = x2(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
..................................................
.....................................................(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)
.............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
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