Linear approximations. Find x so that the linear approx is within 0.1

Question

Verify the given linear approximation at $a = 0$. Then determine the values of x for which the linear approximation is accurate to within 0.1.

So first I verified that the linear approximation is right.

$$e^x \cdot \cos x \approx 1 + x$$

Check $$f'(x) = e^x \cdot (-\sin{x}) + \cos{x} \cdot e^x$$ $$\text{linear approx} = e^x \cdot \cos{x} + (e^x \cdot (-\sin{x}) + \cos{x} \cdot e^x)(x-a)$$

$$L(0) = 1 + 1(x-0) = 1+x$$

b. What are the values of x so that the linear approximation is within 0.1?

So I think I setup this equation correctly?

$$-0.1 < e^x\cos{x} - 1 - x < 0.1$$

Where do I go from here?

Answer 1

The next approximation would be the quadratic one, which would be of the form $1+x+\frac 12f''(0)x^2$. It will not be exact, but a very good approximation is to find the range of $x$ for which $\frac 12f''(0)x^2\lt 0.1$, so evaluate $f''(0)$ and plug in.

Answer 2

$f(x) = f(a) + f'(a)(x-a) + \epsilon$

That is we can our function equals this linear approximation plus some error term.

$|\epsilon| \le |\frac {f''(\xi)}{2!}(x-a)^2|$

In this case $a= 0$ and $f''(x) = -2e^x\sin x$

We want to find the largest $x$ such that

$|\frac {-2e^x\sin x}{2} x^2| < 0.1$

We are not going to find that explicitly but $\frac {2e^x\sin x}{2} \approx x^2$

$x^4 < 0.1 \implies x=0.56$ would be a good guess.

$f(0.56) \approx 1.483$ and our linear approximation is $1.56$ so we are within $0.08$

By trial and error, I find $f(0.608)$ at the point where we break tollerance.