Question

Find the linear approximation of the function g(x) = 3 1 + x at a = 0. g(x) ≈ Correct: Your answer is correct. Use it to approximate the numbers 3 0.95 and 3 1.1 . (Round your answers to three decimal places.)

Answer 1

Answer:

$3.296x+3$

2.835

3.330

Step-by-step explanation:

$g(x) = 3^{1+x}$

Let the linear approximation be L(x) at a = 0. This is given by

$L(x) = g(a) + g'(a)(x-a)$

$g'(x)$ is the derivative of $g(x)$. To find this, we use the form

If $f(x) = a^x$, then $f'(x) =a^x\ln a$

By doing this and applying chain rule for the power (which is a function of x), we have

$g'(x) = 3^{1+x}\ln 3$

Then

$g'(0) = 3^{1+0}\ln 3 = 3.296$

Also $g(0) = 3^{1+0} = 3$

Hence $L(x) = 3+(x-0)\times3.296 = 3.296x + 3$

For $3^{0.95}$, $1+x = 0.95$ and $x=-0.05$

Using this in $L(x)$,

$3^{0.95} = 3.296(-0.05) + 3 = 2.835$

For $3^{1.1}$, $1+x = 1.1$ and $x=0.1$

Using this in $L(x)$,

$3^{1.1} = 3.296(0.1) + 3 = 3.330$