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
1 answer:
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
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Answer:
$1,179
Step-by-step explanation:
Lets use the compound interest formula provided to solve this:
<em>P = initial balance</em>
<em>r = interest rate (decimal)</em>
<em>n = number of times compounded annually</em>
<em>t = time</em>
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First, lets change 2.6% into a decimal:
2.6% -> -> 0.026
Since the interest is compounded quarterly, we will use 4 for n. Lets plug in the values now:
The account balance after 10 years will be $1,179