Find the linear approximation of the function g(x) = 5 1 + x at a = 0. g(x) ≈ use it to approximate the numbers 5 0.95 and 5 1.1 . (round your answers to three decimal places.) 5 0.95 ≈ 5 1.1 ≈ illustrate by graphing g and the tangent line. webassign plot webassign plot webassign plot webassign plot

Differentiating the function

... g(x) = 5^(1+x)

we get

... g'(x) = ln(5)·5^(1+x)

Then the linear approximation near x=0 is

... y = g'(0)(x - 0) + g(0)

... y = 5·ln(5)·x + 5

With numbers filled in, this is

... y ≈ 8.047x + 5 . . . . . linear approximation to g(x)

Using this to find approximate values for 5^0.95 and 5^1.1, we can fill in x=-0.05 and x=0.1 to get

... 5^0.95 ≈ 8.047·(-0.05) +5 ≈ 4.598 . . . . approximation to 5^0.95

... 5^1.1 ≈ 8.047·0.1 +5 ≈ 5.805 . . . . approximation to 5^1.1

not sure but it may be b, as i said i'm not sure.