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
566353621818281727.1232321.2.23121232132312>3454535435454/.435
B hope this helps good luck
http://goblues.org/faculty/grantm/files/2015/12/Practice-Test-3.7-Solutions.pdf
I think it's question number three. Sorry if the link dosen't work.
<h3>
Answer: -1.5</h3>
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Explanation:
Draw a horizontal line from W to the y axis. You should arrive somewhere between -1 and -2. While this isn't exact, it looks like we should arrive right at the middle of -1 and -2; therefore we should get to -1.5
The y coordinate of W is -1.5
Keep in mind this is based on the assumption we reach the halfway point. Unfortunately, W is not on any horizontal grid lines to be able to determine exactly where W is along the y axis.
