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.
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?
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!!
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!!!
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!!!
To determine the instantaneous rate of change at a given point, you need to find the derivative of the function. The derivative measures the rate at which the function is changing with respect to its independent variable (in this case, x). Once you have the derivative, you can plug in the specific value of x to calculate the instantaneous rate of change.
A. To find the instantaneous rate of change at the point where x=2, you need to find the derivative of the function f(x) = x^3 - 7x^2 + 6x - 1. The derivative of f(x) can be found by applying the power rule to each term of the function. The power rule states that the derivative of x^n is n*x^(n-1).
So, let's find the derivative of f(x):
f'(x) = 3x^2 - 14x + 6
Now, substitute x=2 into the derivative equation:
f'(2) = 3(2)^2 - 14(2) + 6 = 12 - 28 + 6 = -10
Therefore, the instantaneous rate of change at the point where x=2 is -10.
B. Similarly, to find the instantaneous rate of change at the point where x=6, use the derivative of f(x):
f'(x) = 3x^2 - 14x + 6
Substitute x=6 into the derivative equation:
f'(6) = 3(6)^2 - 14(6) + 6 = 108 - 84 + 6 = 30
Therefore, the instantaneous rate of change at the point where x=6 is 30.
C. To determine the special kind of point between the points A and B, you need to analyze the behavior of the derivative. Specifically, look for places where the derivative equals zero or is undefined. These points are called critical points.
In this case, find the x-values where f'(x) = 0 or is undefined. To find the critical points, set the derivative equation equal to zero and solve for x:
3x^2 - 14x + 6 = 0
Unfortunately, this is a quadratic equation, and its solutions can be quite complex. You might need to use the quadratic formula to find the exact values of the critical points.
Once you have the critical points, you can analyze the behavior of the original function f(x) to determine whether these points represent local maximums, local minimums, or other special cases.