Question

find the linear approximation of function of g(x)= (1+x)^1/3 at
a=0 use it to approximation(95)^1/3 and (1.1)^1/3​

Answer 1

Step-by-step explanation:

I hope this will help you.

Answer 2

We have,

$g(a) = g(0) = {(1 + 0)}^{ \frac{1}{3} } = 1$

Now taking the derivative.

$g'(x) = \frac{1}{3} (1 + x {)}^{ - \frac{2}{3} }$

$g'(a) = g'(0) = \frac{1}{3} (1 + 0 {)}^{ - \frac{2}{3} } = \frac{1}{3}$

Now we find the equation of the tangent.

$y</p><p>−</p><p>y</p><p>1</p><p>=</p><p>m</p><p>(</p><p>x</p><p>−</p><p>x</p><p>1</p><p>)$

$y</p><p>−</p><p>1</p><p>=</p><p> \frac{1}{3} </p><p>(</p><p>x</p><p>−</p><p>0</p><p>)</p><p>$

$y = \frac{1}{3} x + 1$

As you can see this approximates the function relatively well for value of a in the region of 0.