I am so confused! The problem is:
Verify the given linear approximation at a=0. Then determine the values of x for which the linear approximation in accurate to within 0.1.
LN(1+x) ≈ x
I found f(0) = 0 and f'(0) = 1
I have an equation:
LN (1+x) -(.1) < x < LN(1+x) + 1
The answer is -.383 < x < .516
How did they get those numbers? I know that the line I found approximates the graph of LN(x+1), but I don't know how to determine accuracy to .1 . Thanks for any help - Mary
To verify the given linear approximation at a=0, we need to find the equation of the tangent line to the graph of f(x) = LN(1+x) at x = 0.
First, find the value of f(0):
f(0) = LN(1+0) = LN(1) = 0
Next, find the value of f'(x):
f'(x) = d/dx [LN(1+x)]
Using the chain rule and the fact that the derivative of LN(x) is 1/x:
f'(x) = 1 / (1+x)
Evaluate f'(x) at x = 0:
f'(0) = 1 / (1+0) = 1
So, we have f(0) = 0 and f'(0) = 1. This means that the equation of the tangent line to the graph of f(x) at x = 0 is given by:
y = f'(0)(x-0) + f(0)
y = x + 0
y = x
Now, to determine the values of x for which the linear approximation is accurate to within 0.1, we need to consider the interval around x=0 where the difference between the linear approximation and the actual function value is less than 0.1.
The linear approximation is given by LN(1+x) ≈ x.
So, for the linear approximation to be accurate to within 0.1, we need:
|LN(1+x) - x| ≤ 0.1
Simplifying this inequality, we get:
-0.1 ≤ LN(1+x) - x ≤ 0.1
To determine the values of x that satisfy this inequality, we solve it step by step:
1. LN(1+x) - x ≥ -0.1
This gives us: LN(1+x) ≥ x - 0.1
2. LN(1+x) - x ≤ 0.1
This gives us: LN(1+x) ≤ x + 0.1
Now, solving these inequalities separately:
1. LN(1+x) ≥ x - 0.1:
Since LN(1+x) is an increasing function, we can exponentiate both sides to get rid of the logarithm:
1+x ≥ e^(x-0.1)
To solve this equation, isolate the variable x:
x ≥ e^(x-0.1) - 1
Since we are only interested in x around 0, we evaluate:
0 ≥ e^(-0.1) - 1
0 ≥ 1.105 - 1
0 ≥ 0.105
So, the first part of the inequality is always satisfied for x around 0.
2. LN(1+x) ≤ x + 0.1:
Since LN(1+x) is a concave-down function, we can exponentiate both sides to eliminate the logarithm:
1+x ≤ e^(x+0.1)
To solve this equation, isolate the variable x:
x ≤ e^(x+0.1) - 1
Using a numerical method to solve this, we find that x ≤ 0.516.
Combining the results, we have -0.1 ≤ LN(1+x) - x ≤ 0.1 if -0.1 ≤ x ≤ 0.516.
So, the values of x for which the linear approximation LN(1+x) ≈ x is accurate to within 0.1 are: -0.1 ≤ x ≤ 0.516, which matches the answer you provided: -.383 < x < .516.