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

7. d. 64

8. i. 32

9. c. 125

10. j. none of the above

11. d. 81

12. i. 216

13. e. none of the above

14. g. 240

15. e. none of the above

16. g. 32

17.   e. none of the above