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estimate the instantaneous rate of change of f at x=4

x 3 3.5 4 4.5 5 5.5 6 f(x) 10 8 7 4 2 0 -1  

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Finding the instantaneous rate of change of the function $f(x ...


Finding the instantaneous rate of change of the function $f(x)=-x^2+4x$ at $x=5$, I know the formula for instantaneous rate of change is $\frac{f(a+h)-f(a)}{h}$ I ...
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Instantaneous Rate of Change Formula | [email protected]


The Instantaneous Rate of Change Formula provided with limit exists is, ... Question 1: Calculate the Instantaneous rate of change of the function f(x) ...
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Section 9.3, Average and Instantaneous Rates of Change: The ...


Section 9.3, Average and Instantaneous Rates of Change: The Derivative 1 Average Rate of Change The average rate of change of a function y = f(x) from x = a to x = b is:
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Instantaneous Rate of Change: The Derivative


2 Instantaneous Rate of Change: The Derivative 2.1 The slope of a function Suppose that y is a function of x, say y = f(x). It is often necessary to know how sensitive
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Estimate the instantaneous rate of change of f(x)= x/ (x-4 ...


Answer to Estimate the instantaneous rate of change of f(x)= x/ (x-4) at the point (2, 21).
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Instantaneous Rate of Change


A logical extrapolation would indicate that the instantaneous rate of change in F(x) at a ... directly calculate the slope of the tangent line.
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Instantaneous Rate of Change at a Point - Calculus | Socratic


instantaneous rate of change is like the speed ... How do you estimate instantaneous rate of change at ... How do you find the instantaneous slope of #y=4# at x ...
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Instantaneous Rate of Change — Lecture 8. The Derivative.


Instantaneous Rate of Change — Lecture 8. The Derivative. Recall that the average rate of change of a function y = f(x) on an interval from x 1 to x
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2.1 Instantaneous Rate of Change - Arkansas Tech University


approaches what we would intuitively call the instantaneous velocity at ... The instantaneous rate of change of a function f(x) ... we can estimate f0(5) ...
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Urgent help, What is answer to this rate of change question ?

I read s(x)=60-40e^(-0.05x), where x is advertising cost. I read that s is sales in thousands of dollars. Let's look at the table. When x=14.5, 1000s=60-40e^-0.725=$40627; 1000s(16)=$42027; 1000s(18)=$43737. In each case the result shown is amount in dollars, rather than thousands of dollars, hence 1000s. The rate of change is given by ds/dx=40*0.05e^(-0.05x)=2e^(-0.05x). This is the rate of change of sales revenue with increasing advertising cost. The rate of change per year is ds/dt where t is time in years. ds/dt=2e^(-0.05x)dx/dt. In other words the rate of change in sales revenue per year is related to the rate of change of advertising costs per year. The table has two changes: first is change between years 1 and 2 (1.5); second is change between years 2 and 3 (2). The corresponding change in sales is $1400 and $1710. The average change in advertising costs is 1.75 thousands of dollars; the average change in sales is $1555. The derivative for each of the three years is $968, $899 and $813, giving an average of $893.   More...
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estimate the instantaneous rate of change of f at x=4

at x=3.5, f=8 at x=4.5, f=4 deltax=1, deltaf=-4 rate =deltaf/deltax=-4
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For the function f(x)=x^2-4x:

a. The rate of change is given by f'(x)=2x-4. The average rate of change if (f'(4)+f'(1))/2=(4-2)/2=1. 2x-4 is a linear function, and if we take integer values of x and average them we get: (-2+0+2+4)/4=1. If we increment x by 0.5 steps we get: (-2-1+0+1+2+3+4)/7=1, and so on. b. Instantaneous change of direction at x=1 is -2.
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area of a square with side length x is x2. Instantaneous rate of change in area if x is 5

area of a square with side length x is x^2. Instantaneous rate of change in area if x is 5 We can write the area A as, A = x^2 where the area, A, is a function of x. The rate of chabge of A is dA/dx, where dA/dx = 2x When x = 5, then dA/dx = 2*5 = 10 Answer: Rate of change of area of square, when side length x = 5, is 10
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Estimate the instantaneous rate of change using M(4.9) and (5.1)

M(t) = 1000(1.1)^t M(t) = 1000 * e^((ln1.1)t) M'(t) = 1000(ln1.1) e^((ln1.1)t) M'(t) = 1000(ln1.1)(1.1)^t   M'(4.9) = 1000(ln1.1)(1.1)^(4.9) = 152.0419553 M'(5.1) = 1000(ln1.1)(1.1)^(5.1) = 154.967984
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I need help for this problem. Thanks. I really appreciate the help. How do I find theta?

The minute hand moves 12 times as fast as the hour hand, because the hour hand moves 1/12 of the area of the clock face while the minute hand moves the whole area of the clock face. At 4 o'clock, the hands are 120 degrees apart (one third of 360 degrees). The tips of the hands at this time are distance x apart where x is given by the cosine rule: x=sqrt(3^2+4^2-2*3*4cos120)=sqrt(25+12)=sqrt(37), because cos120=-1/2. At time t minutes after 4, the angle between the tips of the hands decreases initially. For example, if t=15 minutes, the hour hand will move 1/4 the distance between 4 and 5, i.e., 30/4=7.5 degrees, while the minute will have moved 90 degrees. The angle between the hands becomes 120-90+7.5=37.5 degrees. For t minutes, then, the angle ("theta") changes to 120-360t/60+30t/60 degrees=120-6t+t/2=120-11t/2. (Theta can be found for any time t minutes after 4 o'clock from this formula. Also, t can be found when theta is zero, the hands being aligned: t=240/11.) So, x=sqrt(25-24cos(120-11t/2))=sqrt(25-24(cos120cos(11t/2)+sin120sin(11t/2)). If t is very small cos(11t/2) is close to 1 and sin(11t/2) is close to 11t/2. sqrt(120)=sqrt(3)/2; x=sqrt(25-24(-1/2+11tsqrt(3)/4))=sqrt(37-264tsqrt(3)/4)=sqrt(37-66tsqrt(3)). The change in x is sqrt(37-66tsqrt(3))-sqrt(37)=sqrt(37)(sqrt(1-66tsqrt(3)/37)-1). We can use the Binomial Theorem to evaluate sqrt(1-66tsqrt(3)/37)=(1-66tsqrt(3)/37)^(1/2). When t is very small we can ignore t^2 terms and higher powers of t, so we get 1-33tsqrt(3)/37 as an approximation. The change in x becomes sqrt(37)(1-33tsqrt(3)/37-1)=-33tsqrt(3/37). The instantaneous rate of change of x at 4 o'clock is found by dividing this expression by t=-33sqrt(3/37)=-9.3967"/minute=-0.1566"/sec. The minus sign indicates that the hands are closing at the rate of 0.1566 inches per second. The solution can also be found applying formal calculus, where dx (the change in x) is related to the angle change (dø) and the angular rate of change dø/dt can be related to the rate of change of x, dx/dt. This involves differentiating the cosine expression. Because a time of 4 o'clock has been specified, the solution has been calculated from first principles, thus avoiding calculus.  
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what is the formula for calculating the instantaneous rate of change for f(x)=e^x where x=2

d/dx av e^x=e^x...1 av the basik propertees av e e^x at x=2=e^2 or about 7.389046
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Find the instantaneous rate of change of the curve f(x)=(x)/(2-3x) at x=2 using first principles


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cost of equity


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Math for Business

Lechmere Corporation buys word processors from a wholesaler.  The word processors have a $425 list price with a 40% trade discount.  What is the trade discount amount?  What is the net price of the word processor?  Freight charges are  FOB destination.  
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