Find the linear approximation of f(x)=ln x at x=1 and use it to estimate ln 1.12.
L(x)= ?
ln 1.12 \approx ?
f' = 1/x
f'(1) = 1
f(1) = 0
So, you have a point (1,0) and a slope (1), so
y = x-1
now plug in x = 1.12
To find the linear approximation of f(x) = ln(x) at x = 1, we can use the formula:
L(x) = f(a) + f'(a)(x - a)
where a is the value at which we are finding the linear approximation (x = 1 in this case), f(a) is the value of the function at a, and f'(a) is the derivative of the function evaluated at a.
First, let's find f(a) and f'(a) for f(x) = ln(x) at x = 1:
- f(a) = ln(1) = 0
- f'(a) = 1/a = 1/1 = 1
Now, we can find L(x) using the formula:
L(x) = f(a) + f'(a)(x - a)
L(x) = 0 + 1(x - 1)
L(x) = x - 1
So, L(x) = x - 1 is the linear approximation of f(x) = ln(x) at x = 1.
Now, let's use this linear approximation to estimate ln(1.12). Since a = 1, we substitute x = 1.12 into L(x) = x - 1:
LN(1.12) ≈ (1.12) - 1
≈ 0.12
Therefore, ln(1.12) is approximately 0.12 using the linear approximation.
To find the linear approximation of f(x)=ln x at x=1, we can use the formula for the linear approximation:
L(x) = f(a) + f'(a)(x - a)
where a is the value at which we are finding the linear approximation, f(a) is the value of the function at that point, and f'(a) is the derivative of the function at that point.
First, let's find f(a) and f'(a) for f(x)=ln x at x=1.
f(a) = f(1) = ln(1) = 0
f'(a) = f'(1) = 1/x at x=1 = 1/1 = 1
Now we can plug these values into the formula:
L(x) = 0 + 1(x - 1) = x - 1
So, the linear approximation of f(x)=ln x at x=1 is L(x) = x - 1.
To estimate ln 1.12 using this linear approximation, we can substitute x = 1.12 into the linear approximation:
L(1.12) = 1.12 - 1 = 0.12
Therefore, ln 1.12 is approximately 0.12.