# a) Use a linear approximation to estimate 1/(103). b) Use the Intermediate Value Theorem (IVT) to...

## Question:

a) Use a linear approximation to estimate {eq}\displaystyle \dfrac 1{103} {/eq}.

b) Use the Intermediate Value Theorem (IVT) to show that {eq}\sin (x) = x - 1 {/eq} has a solution in the interval {eq}\displaystyle [0,\ \pi] {/eq}.

## Linear Approximation:

In order to estimate, use taylor expansion {eq}(1+x)^{-a}=1+ax+O(x^2) {/eq}

As per intermediate value theorem, if afunction f(x) is continuous on the interval a,b such that it has values of opposite sign inside this interval, then there must be some value x = d on the interval (a,b) for which f(d)=0.

a)

{eq}\frac{1}{103}=\frac{1}{100\left ( 1+\frac{3}{100} \right )}\\ =\frac{1}{100}\left ( \frac{1}{1+x} \right )\,,\,x=\frac{3}{100}<1 {/eq}

As per taylor's expansion, {eq}(1+x)^{-a}=1+ax+O(x^2) {/eq}

So,

{eq}\frac{1}{103}=\frac{1}{100\left ( 1+\frac{3}{100} \right )}\\ =\frac{1}{100}\left ( \frac{1}{1+x} \right )\,,\,x=\frac{3}{100}<1\\ =\frac{1}{100}\left ( 1-\frac{3}{100}+O(x^2) \right )\\ \approx \frac{1}{100}\left ( 1-\frac{3}{100} \right )\\ =\frac{1}{100}\left ( \frac{97}{100} \right )\\ =0.0097 {/eq}

b)

Take {eq}f(x)=\sin x-x+1 {/eq}

At {eq}x=0\,,\,\pi {/eq},

{eq}f(0)=\sin 0-0+1=1>0\\ f(\pi)=\sin \pi-\pi+1=0-\pi+1=-3.14+1=-2.14<0 {/eq}

So, as per intermediate value theorem, f has atleast one solution in the interval {eq}\displaystyle [0,\ \pi] {/eq}