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avatar+771 

Consider the function f(x)=2x3-3x

 

a) Find the average rate of change between the points on the function when x=1 and x=2.

 

b) On a graph, a line that connects any two points is called a secant line. Find the equation of the secant line that connects the points on the function when x=1 and x=2.

 

c) In calculus, we find the slope between a point on the graph when x=a and another point that is a small distance away. We use x=a+h for the second point. Find the average rate of change between the points x=a and x=a+h.

 Oct 25, 2017

Best Answer 

 #3
avatar+7352 
+2

a)     average rate of change  =

 

 \(\frac{\text{change in f(x)}}{\text{change in x}}\,=\,\frac{f(1)-f(2)}{1-2}\,=\,\frac{(\,2(1)^3-3(1)\,)\,-\,(\,2(2)^3-3(2)\,)}{1-2}\,=\,\frac{(-1)-(10)}{-1}\,=\,\frac{-11}{-1}\,=\,11\)

 

like helperid said.

 

b)     We want an equation of a line that has a slope of  11  and passes through  the point (1, f(1) )

 

And  (1, f(1)  =  (1, -1)

 

So..   the equation in point-slope form is     y + 1  =  11(x - 1)

And in slope intercept form, it is   y =  11x - 12

 

c)     Find the average rate of change between the points  x = a  and  x = a + h .

 

Thie is the same as part  a  , just with letters in place of numbers.   laugh

 

average rate of  change  =

 

\(=\,\frac{\text{change in f(x)}}{\text{change in x}}\\~\\ =\,\frac{f(a+h)-f(a)}{(a+h)-(a)} \\~\\ =\,\frac{(\,2(a+h)^3-3(a+h)\,)\,-\,(\,2(a)^3-3(a)\,)}{h} \\~\\ =\,\frac{2(a+h)^3-3(a+h)-2a^3+3a}{h} \)

 

I don't know if you need it more simplified than that.

 Oct 25, 2017
 #1
avatar+556 
+3

a) 11

b) g(x)=11x - 12

c) idk

 Oct 25, 2017
 #2
avatar+771 
+3

Could you explain how to do these though? Just the answers does me no good.

AdamTaurus  Oct 25, 2017
 #3
avatar+7352 
+2
Best Answer

a)     average rate of change  =

 

 \(\frac{\text{change in f(x)}}{\text{change in x}}\,=\,\frac{f(1)-f(2)}{1-2}\,=\,\frac{(\,2(1)^3-3(1)\,)\,-\,(\,2(2)^3-3(2)\,)}{1-2}\,=\,\frac{(-1)-(10)}{-1}\,=\,\frac{-11}{-1}\,=\,11\)

 

like helperid said.

 

b)     We want an equation of a line that has a slope of  11  and passes through  the point (1, f(1) )

 

And  (1, f(1)  =  (1, -1)

 

So..   the equation in point-slope form is     y + 1  =  11(x - 1)

And in slope intercept form, it is   y =  11x - 12

 

c)     Find the average rate of change between the points  x = a  and  x = a + h .

 

Thie is the same as part  a  , just with letters in place of numbers.   laugh

 

average rate of  change  =

 

\(=\,\frac{\text{change in f(x)}}{\text{change in x}}\\~\\ =\,\frac{f(a+h)-f(a)}{(a+h)-(a)} \\~\\ =\,\frac{(\,2(a+h)^3-3(a+h)\,)\,-\,(\,2(a)^3-3(a)\,)}{h} \\~\\ =\,\frac{2(a+h)^3-3(a+h)-2a^3+3a}{h} \)

 

I don't know if you need it more simplified than that.

hectictar Oct 25, 2017
 #4
avatar+771 
+3

That's good enough. Thanks Hectictar!

laugh

AdamTaurus  Oct 25, 2017

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