This appears to be from an intro to calculus course.
"instantaneous rate of change" is equivalent to slope, so the questions A and B become "what is the slope of the function" at these points.
Have you done differentiation? If you have, you should be able to read off the slope at these points.
If you haven't, I presume your course must have covered some way of approximating or calculating the slope at a point.
As for C: at 2, the slope is negative; at 6, it's positive. Assuming the change in slope is continuous, it must have passed though zero at least once in the interval. Does this ring a bell?
Consider the function f(x)=x^3-7X^2+6x-1
A. Determine the instantaneous rate of change at the point where x=2
B. Determine the instantaneous rate of change at the point where x=6
C. What special kind of point occurs between the points A and B? Explain.
I don't understand this at all. What am I supposed to do. Should I graph it? Or is there an algebraic way to do it. I hav been trying to figure out the solution for a qhile and still, nothing. Please help me.
4 answers
You can do this by 3 methods.(known to me)
1. Draw a table, that helps you approximate the instantaneous rate of change by using the idea of average rate of change.( bring 'h', the difference of 2 close points)
2. Drawing a graph and finding the slope of the tangent line at points; x=2 and x=6.
3. Using Difference Quotient, where instantaneous rate of change is determined by the formula;
[f(a+h) –f(a)]/h
A. Determine the instantaneous rate of change at the point where x=2
[f(a+h) –f(a)]/h
Let a=2(same as ‘x’), and h=0.01(a value of your choice)
Hence, [f(2.01)-f(2)]/0.01
F(2.01) = -9.100099
F(2) = -9
Therefore; [-9.100099-(-9)]/0.01
= -10.0099
B. Do the same for x=6, here a=6!!
C. I assume the question is asking for the average rate of change.
Hoping the above mentioned statements are right and that it may be of some help to you!!
-Cheers. Have a great day!!
1. Draw a table, that helps you approximate the instantaneous rate of change by using the idea of average rate of change.( bring 'h', the difference of 2 close points)
2. Drawing a graph and finding the slope of the tangent line at points; x=2 and x=6.
3. Using Difference Quotient, where instantaneous rate of change is determined by the formula;
[f(a+h) –f(a)]/h
A. Determine the instantaneous rate of change at the point where x=2
[f(a+h) –f(a)]/h
Let a=2(same as ‘x’), and h=0.01(a value of your choice)
Hence, [f(2.01)-f(2)]/0.01
F(2.01) = -9.100099
F(2) = -9
Therefore; [-9.100099-(-9)]/0.01
= -10.0099
B. Do the same for x=6, here a=6!!
C. I assume the question is asking for the average rate of change.
Hoping the above mentioned statements are right and that it may be of some help to you!!
-Cheers. Have a great day!!
A. Determine the instantaneous rate of change at the point where x=2
Let a=2(same as ‘x’), and h=0.01(a value of your choice)
Hence, [f(2.01)-f(2)]/0.01
F(2.01) = -9.100099
F(2) = -9
Therefore; [-9.100099-(-9)]/0.01
= -10.0099
B. Do the same as above, except that the x=2 must be replaced with 6. (Therefore, a=6)
C. I assume that the question asks for average rate of change. (The points that fall between the points might be on the secant lines.)
Hoping that the above statements are correct and that the solution may be of some help to you!!
-Cheers! Have a great day!!!
Let a=2(same as ‘x’), and h=0.01(a value of your choice)
Hence, [f(2.01)-f(2)]/0.01
F(2.01) = -9.100099
F(2) = -9
Therefore; [-9.100099-(-9)]/0.01
= -10.0099
B. Do the same as above, except that the x=2 must be replaced with 6. (Therefore, a=6)
C. I assume that the question asks for average rate of change. (The points that fall between the points might be on the secant lines.)
Hoping that the above statements are correct and that the solution may be of some help to you!!
-Cheers! Have a great day!!!
Determine the instantaneous rate of change at the point where x=2
Let a=2(same as ‘x’), and h=0.01(a value of your choice)
Hence, [f(2.01)-f(2)]/0.01
F(2.01) = -9.100099
F(2) = -9
Therefore; [-9.100099-(-9)]/0.01
= -10.0099
Do the same as above for B., except that the x=2 must be replaced with 6. (Therefore, a=6)
I assume that the question C. asks for average rate of change. (The points that fall between the points might be on the secant lines.)
Hoping that the above statements are correct and that the solution may be of some help to you!!
-Cheers! Have a great day!!!
Let a=2(same as ‘x’), and h=0.01(a value of your choice)
Hence, [f(2.01)-f(2)]/0.01
F(2.01) = -9.100099
F(2) = -9
Therefore; [-9.100099-(-9)]/0.01
= -10.0099
Do the same as above for B., except that the x=2 must be replaced with 6. (Therefore, a=6)
I assume that the question C. asks for average rate of change. (The points that fall between the points might be on the secant lines.)
Hoping that the above statements are correct and that the solution may be of some help to you!!
-Cheers! Have a great day!!!