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# two problems that i'm seriously stuck on. any help!

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1769
2
+31

1. The table of values represents a reciprocal function ​f(x).

How much greater is the average rate of change over the interval [−6, −4][−6, −4] than the interval [−3, −1][−3, −1] ?

x                     f(x)

−6                  −0.0046

−5                  −0.0079

−4                  −0.0154

−3                  −0.0357

−2                  −0.1111

−1                  −0.5

2. Consider two functions: g(x)=20(1.5)x and the function ​f(x)​ shown in the table.

Which statements are true?

x              f(x)

−5           −45

−4           −48

−3           −49

−2           −48

−1           −45

0             −40

1             −33

a) ​g(x)​ has a greater y-intercept than ​f(x)​ does.

b) ​f(x)​ increases at a faster rate than ​g(x)​ does on the interval (−5, −3) .

c)​ f(1) is less than g(−1) .

d) f(x) and ​g(x)​ are both increasing on the interval  (−∞, ∞) .

Feb 5, 2018

#1
+7598
+3

1.    Here's a graph of the points to get an idea of what the function looks like.

average rate of change   =   $$\frac{\text{change in }f(x)}{\text{change in }x}$$

average rate of change over the interval  [-6, -4]   =   $$\frac{ f(-6) \,-\, f(-4)}{ (-6) \,-\,( -4) }$$

$$\frac{ f(-6) \,-\, f(-4)}{ (-6) \,-\,( -4) }\,=\,\frac{ (-0.0046) \,-\, (-0.0154) }{ (-6) \,-\, (-4) }\,=\,\frac{ 0.0108 }{-2 }\,=\,- 0.0054$$

average rate of change over the interval  [-6, -4]   =   - 0.0054

Notice that this is just the slope of the line through the points  ( -6, f(-6) )  and  ( -4, f(-4) ) .

average rate of change over the interval  [-3, -1]   =   $$\frac{ f(-3) \,-\, f(-1)}{ (-3) \,-\,( -1) }$$

$$\frac{ f(-3) \,-\, f(-1)}{ (-3) \,-\,( -1) }\,=\, \frac{ (-0.0357) \,-\, (-0.5)}{ (-3) \,-\,( -1) }\,=\, \frac{0.4643}{ -2 }\,=\,- 0.23215$$

average rate of change over the interval  [-3, -1]   =   - 0.23215

How much greater is  - 0.0054  than  - 0.23215  ?

(-0.0054)  -  (-0.23215)   =   0.22675

Feb 5, 2018
#2
+7598
+2

2.     Assuming that     g(x)   =   20(1.5)x

the y-intercept of g(x)  =   g(0)   =   20(1.5)0   =   20(1)   =   20

the y-intercept of f(x)  =  f(0)   =   -40

The y-intercept of  g(x)  is greater than the y-intercept of f(x) , so  a)  is true.

From  -5  to  -3 ,  f(x)  goes from  -45  to  -49 ,  and   (-49) - (-45)   =   -4

From  -5  to  -3 ,  g(x)  goes from  $$\frac{640}{243}$$  to  $$\frac{160}{27}$$ , and  $$\frac{160}{27}$$  -  $$\frac{640}{243}$$   =   $$\frac{800}{243}$$

f(x)  is decreasing on the interval  (-5, -3)  because as  x  gets larger, f(x) gets smaller.

$$\frac{800}{243}$$  >  -4 ,  so  g(x)  is increasing faster than  f(x)  on the interval  (-5, -3) .  b)  is false.

f(1)   =   -33

g(-1)   =   20(1.5)-1   =   $$\frac{20}{1.5}$$   =   $$\frac{40}{3}$$   ≈   13.3

Is  f(1)  less than  g(-1) ?   Is  -33  less than  13.3 ?  Yes, so  c)  is true.

We already found that  f(x)  is decreasing on the interval  (-5, -3)  , so

f(x)  can't be increasing on the interval  (−∞, ∞) .  d)  is false.

Feb 6, 2018