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
To determine the values of x for which the linear approximation is accurate to within 0.1, you will need to use the equation you found:
LN(1+x) - 0.1 < x < LN(1+x) + 0.1
First, let's simplify this equation:
LN(1+x) - 0.1 < x < LN(1+x) + 0.1
Next, let's substitute the Taylor series approximation for LN(1+x) which is x:
x - 0.1 < x < x + 0.1
Simplifying this further, we can cancel out the x terms:
-0.1 < 0 < 0.1
As you can see, the inequality -0.1 < 0 < 0.1 is always true for any value of x. Therefore, the linear approximation is accurate to within 0.1 for all values of x.
Now, to clarify your statement about the range of x as -.383 < x < .516, it seems like there might be a mistake in the calculations. The range of x for which the linear approximation is accurate to within 0.1 is actually the entire real number line, not a specific interval.
I hope this clears up any confusion. Let me know if you have any further questions!
To determine the values of x for which the linear approximation is accurate to within 0.1, we can start by understanding the linear approximation itself.
The linear approximation of a function f(x) at a specific point a is given by the equation:
f(x) ≈ f(a) + f'(a)(x - a)
In this case, the given function is LN(1+x), and we want to verify the linear approximation at a = 0, which means f(a) = LN(1+0) = LN(1) = 0.
We also found f'(0) = 1, so the linear approximation becomes:
LN(1+x) ≈ 0 + 1(x - 0)
LN(1+x) ≈ x
Now, we want to find the values of x for which the linear approximation is accurate to within 0.1. This means that we are looking for x values that satisfy the inequality:
-0.1 < LN(1+x) - x < 0.1
To solve this inequality, we need to isolate x. Let's simplify the inequality step-by-step:
1. Start with the inequality: -0.1 < LN(1+x) - x < 0.1
2. Add x to all sides: -0.1 + x < LN(1+x) - x + x < 0.1 + x
3. Simplify the middle term: -0.1 + x < LN(1+x) < 0.1 + x
4. Since LN(1+x) is an increasing function for x > -1, we can exponentiate both sides using the base e (natural logarithm base) to cancel out the logarithm function:
e^(-0.1 + x) < 1+x < e^(0.1 + x)
5. Simplify the exponentiation: e^(-0.1) * e^x < 1+x < e^0.1 * e^x
6. Further simplify: e^(-0.1) * e^x < 1+x < e^0.1 * e^x can be written as:
0.9048e^x < 1+x < 1.1052e^x
Now, we need to solve this inequality for x. Let's break it down into two inequalities:
First inequality: 0.9048e^x < 1+x
1. Subtract x from both sides: 0.9048e^x - x < 1
2. Rearrange terms: 0.9048e^x - x - 1 < 0
3. Solve this inequality for x using numeric or graphical methods. In this case, it is approximately x > -0.383.
Second inequality: 1+x < 1.1052e^x
1. Subtract 1 from both sides: x < 1.1052e^x - 1
2. Rearrange terms: -1 < 1.1052e^x - x
3. Solve this inequality for x using numeric or graphical methods. In this case, it is approximately x < 0.516
Therefore, the values of x for which the linear approximation LN(1+x) ≈ x is accurate to within 0.1 are approximately -0.383 < x < 0.516.