Answer:
[tex]3.296x+3[/tex]
2.835
3.330
Step-by-step explanation:
[tex]g(x) = 3^{1+x}[/tex]
Let the linear approximation be L(x) at a = 0. This is given by
[tex]L(x) = g(a) + g'(a)(x-a)[/tex]
[tex]g'(x)[/tex] is the derivative of [tex]g(x)[/tex]. To find this, we use the form
If [tex]f(x) = a^x[/tex], then [tex]f'(x) =a^x\ln a[/tex]
By doing this and applying chain rule for the power (which is a function of x), we have
[tex]g'(x) = 3^{1+x}\ln 3[/tex]
Then
[tex]g'(0) = 3^{1+0}\ln 3 = 3.296[/tex]
Also [tex]g(0) = 3^{1+0} = 3[/tex]
Hence [tex]L(x) = 3+(x-0)\times3.296 = 3.296x + 3[/tex]
For [tex]3^{0.95}[/tex], [tex]1+x = 0.95[/tex] and [tex]x=-0.05[/tex]
Using this in [tex]L(x)[/tex],
[tex]3^{0.95} = 3.296(-0.05) + 3 = 2.835[/tex]
For [tex]3^{1.1}[/tex], [tex]1+x = 1.1[/tex] and [tex]x=0.1[/tex]
Using this in [tex]L(x)[/tex],
[tex]3^{1.1} = 3.296(0.1) + 3 = 3.330[/tex]
