If f(-1)=-3 and f'(x)=(4x^2)/(x^3+3), which of the following is the best approximation for f(-1.1) using local linearization?
a) -7.2
b) 2.8
c) -1.2
d) -3.2
to approximate f(-1.1), let ∆x = -0.1 and you have
∆y/∆x ≈ dy/dx
∆y ≈ dy/dx * ∆x
dy/dx at x = -1 is (4*(-1)^2))/((-1)^3+3) = 4/2 = 2
so, ∆y ≈ 2(-0.1) = -0.2
thus, y(-1.1) ≈ y(-1) + ∆y = -3 - 0.2 = -3.2
To approximate f(-1.1) using local linearization, we can use the formula:
f(x) ≈ f(c) + f'(c)(x - c)
where c is the value at which we know the function and its derivative, and x is the value we want to approximate.
Given f(-1) = -3, we can use this as our c value. Substituting c = -1 and f'(x) = (4x^2)/(x^3+3) into the formula, we have:
f(-1.1) ≈ f(-1) + f'(-1)(-1.1 - (-1))
f(-1.1) ≈ -3 + (4(-1)^2)/((-1)^3+3) * (-1.1 + 1)
f(-1.1) ≈ -3 + (4 * 1) / (1 + 3) * (-0.1)
f(-1.1) ≈ -3 + 4/4 * (-0.1)
f(-1.1) ≈ -3 + (-0.1)
f(-1.1) ≈ -3 - 0.1
f(-1.1) ≈ -3.1
Therefore, the best approximation for f(-1.1) using local linearization is -3.1.
None of the given options (-7.2, 2.8, -1.2, -3.2) match this approximation, so the correct answer is none of the above.
To approximate the value of f(-1.1) using local linearization, we can use the formula for linear approximation:
L(x) = f(a) + f'(a)(x-a)
where L(x) is the linear approximation of f(x), a is the given point, and x is the desired point to approximate. In this case, a = -1, and we want to approximate f(-1.1).
First, let's substitute the given values into the formula:
L(-1.1) = f(-1) + f'(-1)(-1.1 - (-1))
Now, let's compute the value of f'(-1):
f'(-1) = (4(-1)^2) / (-1^3+3)
= 4 / 2
= 2
Next, substitute the values back into the formula:
L(-1.1) = -3 + 2(-1.1 + 1)
Simplifying further:
L(-1.1) = -3 + 2(-0.1)
= -3 + (-0.2)
= -3 - 0.2
= -3.2
Therefore, the best approximation for f(-1.1) using local linearization is option d) -3.2.