Use linear approximation, i.e the tangent line to approximate 3rd sqrt(1.1) as follows.?
So approximate 3rd sqrt(1.1). Let f(x) = 3rd sqrt(x), the equation of the tangent line f(x)=1 can be written in the form y=mx+b. Find m: Find b: And using this information approximate 3rd sqrt(1.1). Can anyone help me do this. I have found the slope to be 1/3 but I have no idea how to do the last two parts. Please Help. thanks erek
Use linear approximation, i.e. the tangent line, to approximate cube root 64.4 as follows:?
The computer is correct. Evaluate your derivative at x = 64; you should get 1/48. The slope of the tangent line at a given point is equal to the value of the derivative at that point. Then the equation of the tangent line is y - (64^(1/3)) = (1/48)(x - 64) or in slope-intercept form y = (1/48)x + 8/3 To get the approximation, evaluate the tangent line equation at x = 64.4
Use Linear Approximation to approximate the cube root of 125.2?
125 = 5^3 therefor, the cube root for 125 is 5. f(x) = x ^ (1/3) from what i can deduct, m can only be obtained depending on x. and you have: mx+b = x ^ (1/3) so b=0 and mx = x ^ (1/3) which means that m = [x ^ (1/3) ] / x = x^[ (1/3) - 1] = x ^ ( -2/3).
Use linear approximation to approximate cube root of 64.4?
Let y=x^(1/3), that is the cube root function. Then dy/dx = (1/3) x^(-2/3) and thinking of dy and dx as deltay and deltax dy = (1/3) x^(-2/3) dx , approximately. Intuitively this says, that if we go to a particular value of x and increase x by a little bit dx, y will increase from the value of the function at x by the amount dy computed as (1/3) x^(-2/3)dx. So put x = 64, whose cube root is y = 4, and put dx= 0.4 and get (64.4)^(1/3) = 4 + dy = 4 + (1/3) 64^(-2/3)(0.4) = 4 +(1/3)[ 64^(-1/3)]^2 (0.4)= 4 + (1/3)(1/4)^2 (0.4) = 4 + (1/48)(0.4) = 4.0083 approximately.
Use the tangent line approximation (linear approximation) at x=27 to estimate the cube root of 26?
y = x^(1/3) => dy/dx = (1/3)x^(-2/3). In your problem, x = 27, and (1/3)x^(-2/3) = (1/3)*(1/9) = 1/27, so dy = (1/27)dx = (1/27)(-1) = -0.037; the cube root of 26 is near 3 - 0.037 = 2.963. The actual cube root of 26 is near 2.9625.
Using linear approximation, i.e. the tangent line, to approximate sqrt1.2^(1/3)?
I am assuming this is f(x) = x^(1/3) (pronounced as "the cube root of x") rather than f(x) = √[x^(1/3)] = x^(1/6) (which is what you said). First, to find the slope of this tangent line, we need to find f'(x). With the Power Rule: f'(x) = (1/3)x^(1 - 1/3) = (1/3)x^(-2/3) = 1/[3x^(2/3)]. Then, the slope of the tangent line at x = 1 is: f'(1) = 1/[3(1)^(2/3)] = 1/3. So you got that right! Then, we know that the function passes through (1, 1). So we require a line with a slope of 1/3 and passes through (1, 1). With point-slope form, we see that it is: y - 1 = 1/3 * (x - 1) ==> y - 1 = (1/3)x - 1/3 ==> y = (1/3)x + 2/3. So b = 2/3. Then, we see that: x^(1/3) ≈ (1/3)x + 2/3 for x ≈ 1. Thus: (1.2)^(1/3) ≈ (1/3)(1.2) + 2/3 ≈ 1.07. I hope this helps!
Estimate the cubed root of 65 using a tangent line approximation.?
Note that 65^(1/3) is close enough to 64^(1/3) = 4 to use as a good approximation. From this, we can base our approximation using the tangent line to y = x^(1/3) at (64, 4). Since dy/dx = (1/3)x^(1/3 - 1) = 1/[3x^(2/3)], we see that the slope of the tangent line is: m = 1/[3(64)^(2/3)] = 1/[3(16)] = 1/48. Then, since the tangent line passes through (64, 4), the tangent line is: y - 4 = (1/48)(x - 64), by point-slope form ==> y = (1/48)x + 8/3. From this, we can conclude that: x^(1/3) ≈ (1/48)x + 8/3, for x ≈ 64. Thus, 65^(1/3) ≈ (1/48)(65) + 8/3 ≈ 4.0208. I hope this helps!
What is linear approximation and linearization?
I just finished taking Calculus AB last year as a sophomore at Olympian High School, and if I remember correctly, linear approximation is when you use differentials to approximate a certain value that is close to a known value. One example off the top of my head is being asked to approximate ∛26.5.For this question, you must first decide a function that would give you the value of ∛26.5 if you had a table of value for that function. I would chose the function f(x)=∛x, which would follow the above criterion and is also differentiable, which is the most important part of determining which function to use.Next, since you know that if you plug in 27 for x, you'd attain a y-value of 3, which is very easy to manipulate. With this coordinate (27,3) in mind, you must now find the equation of the tangent line when x=2 (this equation will be written in point-slope form). All you need left to write a complete equation for the above-mentioned tangent line is the slope of f(x) when x=27. To find the slope, you must find the first derivative of f(x) and then plug in 27 as the x-value, and whatever value that emerges will be your slope.f(x)=∛26.5. This can be rewritten as f(x)=x^(1/3), which is more easily differentiated. To find the derivative of this function, you need to multiply f(x) by its exponent (1/3) and then subtract that exponent by 1. Thus, the derivative of f(x)=x^(1/3) is〖1/3 x〗^((-2)/3), which, when you plug in 27 as the x-value, yields (1/27). This is your slope.Now, with both a coordinate and a slope, you can write the equation of the tangent line as y-3=(1/27)(x-27). With this equation, you can now complete the final step of this exercise: approximating the value of the cube root of 26.5. This is done by simply substituting in 26.5 for the x-value in the tangent line equation. This substitution yields a y-value of approximately 2.98148.This approximation of 2.98148 is remarkably close the actual value of ∛26.5, which is 2.98136.
Linear approximation... please please help?
I'm assuming in question one that "\sqrt[3]" denotes the cube root function. The primary idea behind linear approximation lies in the fact that that points on the tangent line of a function at the point x are somewhat of a good approximation of the function for values close to x. 1.) In question 1, we are asked to approximate the cube root of 8.3. Note: we know the cube root of 8, which is 2... so we should find the tangent line to f(x) at x = 8. f '(x) = (1/3) x^(-2/3), by the "power" rule, and by the fact that the cube root of x, is x^(1/3). So the slope of the tangent line of f(x) at x = 8, is f '(8) = (1/12). (So m = 1/12, in your question). Also the tangent line contains the point (8,2), since f(8) = 2, and the tangent line passes through the graph at this point. In point-intercept format, the tangent line becomes: (y - 2) = (1/12)(x - 8) y = (1/12)x + 4/3 (so b = 4/3 in your question) This tangent line approximates f(x) fairly well for x close to 8. We use this fact to estimate 8.3. Setting x = 8.3 on our tangent line we get: y = (1/12)8.3 + 4/3 y = (1/12)(83/10) + (4/3) y = (83/120) + (160/120) y = (243/120) = 2.025 And that is our linear approximation. The "real" value for the cube root of 8.3 (at least for 5 decimals) is 2.02469. So the approximation is pretty close to the real value. There is a short cut for solving the above. Finding the equation for the tangent line often takes a few minutes. We can skip that step by noting: the linear approximation of f(x + a) is: f(x + a) = f(x) + a f ' (x) where "=" is approximate In the above case x + a = 8.3 = 8 + 0.3 so f(8.3) = f(8) + (0.3) f ' (8) = 2 + (0.3)(1/12) = 2.025 So the key to these problems is to find a number nearby for which we know the solution (in this case x = 8) and then finding how far we are away from that value (a = 0.3). 2.) f(x) = sqrt{4 + 2x}. We want to find the tangent line for f(x) at x = 0. The slope of the tangent line is m = f ' (0) = sqrt(4) = 2. And a point on the line is (0,f(0)) = (0,2). Hence the equation for the line is: y - 2 = 2(x - 0) y = 2x + 2. So A = 2 and B = 2
Calc - linear approximation?
first, find your tangent line ot the cube root of x d/dx(x^1/3)=(1/3)*x^(-2/3) now let us pick a point we know is close to what we want to estimate, 125 is an easy cube root, so let us pick 125 as the point for our tangent line: f'(125)=(1/3)*(1/25)=1/75 so m, the slope at x=125 is equal to 1/75 now, we also got the initial condition that f(125)=5 so 5=(1/75)*(125)+b y=mx+b so 125/75= 5/3 5=5/3+b so 15=5+3b 10=3b b=10/3 y=(1/75)(x)+10/3 Now plug and chug f(125.2)=(125.2)/(75)+10/3=5.00266667 cube root of 125.2 is about 5.00266667 real cube root is 5.00266525 Look at how close we are : ) Talk to me if you need more explanation. And remember that the slope at a point is the derivative, so m = f'(a) where a is the point at which you want to find the slope.