If f(−1) = −3 and f prime of x equals the quotient of 4 times x squared and the quantity x cubed plus 3 , which of the following is the best approximation for f(–1.1) using local linearization

Question

If f(−1) = −3 and f prime of x equals the quotient of 4 times x squared and the quantity x cubed plus 3 , which of the following is the best approximation for f(–1.1) using local linearization

Answer 1

df/dx = 4 x^2/[x^3 +3]

at x = -1
df/dx = 4/[2] = 2

so
f(-1+dx) = f(-1) + 2 dx
here dx = -.1
so
f(-1-.1) = -3 +2(-.1)
= -3 -.2
=-3.2

Answer 2

Well, let's start by finding the derivative of f(x) using the given information:

f'(x) = (4x^2) / (x^3 + 3)

Now, let's use this derivative to approximate f(-1.1) using local linearization. Since we're using local linearization, we'll approximate f(x) as f(a) + f'(a)(x - a), where a is the value of x we want to approximate around (in this case, a = -1).

f(-1.1) ≈ f(-1) + f'(-1)(-1.1 - -1)

Since we're given that f(-1) = -3, let's plug in the values:

f(-1.1) ≈ -3 + f'(-1)(-0.1)

Now plug in f'(-1) to get the approximation:

f'(-1) = (4(-1)^2) / ((-1)^3 + 3)
= 4 / (-1 + 3)
= 4 / 2
= 2

Plugging this back into the previous equation:

f(-1.1) ≈ -3 + 2(-0.1)
= -3 - 0.2
= -3.2

So, the best approximation for f(-1.1) using local linearization is approximately -3.2.

Answer 3

To approximate the value of f(–1.1) using local linearization, we can use the equation of a line tangent to the graph of f at x = –1.

Step 1: Find the derivative, f'(x):
Given that f'(x) = (4x^2) / (x^3 + 3)

Step 2: Find the slope of the tangent line at x = –1:
Evaluate the derivative at x = –1:
f'(-1) = (4(-1)^2) / ((-1)^3 + 3) = 4 / 2 = 2

Step 3: Write the equation of the tangent line using the point-slope form:
Using the point (-1, f(-1)) = (-1, -3) and the slope m = 2, the equation is:
y - (-3) = 2(x - (-1))

Step 4: Simplify the equation of the tangent line:
y + 3 = 2(x + 1)
y + 3 = 2x + 2
y = 2x - 1

Step 5: Approximate f(-1.1) using the equation of the tangent line:
Plug in x = –1.1 into the equation:
f(-1.1) = 2(-1.1) - 1
f(-1.1) = -2.2 - 1
f(-1.1) ≈ -3.2

Therefore, the best approximation for f(–1.1) using local linearization is f(-1.1) ≈ -3.2.

Answer 4

To approximate f(-1.1) using local linearization, we will use the linear approximation formula:

L(x) = f(a) + f'(a) * (x - a)

Here, "a" is the point around which we are linearizing, and "x" is the value at which we want to approximate the function.

Given that f(-1) = -3, we will consider (-1, -3) as our anchor point.

To find f'(-1), we need to differentiate the function f(x) using the given expression for f'(x):

f'(x) = (4x^2) / (x^3 + 3)

Substituting x = -1 into the expression for f'(x), we get:

f'(-1) = (4 * (-1)^2) / ((-1)^3 + 3)
= 4 / (1 + 3)
= 4 / 4
= 1

So, f'(-1) = 1.

Now, using the linear approximation formula, we substitute a = -1, f(a) = f(-1) = -3, f'(a) = f'(-1) = 1, and x = -1.1:

L(-1.1) = f(-1) + f'(-1) * (-1.1 - (-1))
= -3 + 1 * (-0.1)
= -3 - 0.1
= -3.1

Therefore, the best approximation for f(-1.1) using local linearization is -3.1.