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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4) 6x
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Step-by-step explanation:
We can work both these problems at once by finding an applicable rule.
where O(h²) is the series of terms involving h² and higher powers. When divided by h, each term has h as a multiplier, so the series sums to zero when h approaches zero. Of course, if n < 2, there are no O(h²) terms in the expansion, so that can be ignored.
This can be referred to as the <em>power rule</em>.
Note that for the quadratic f(x) = ax^2 +bx +c, the limit of the sum is the sum of the limits, so this applies to the terms individually:
lim[h→0](f(x+h)-f(x))/h = 2ax +b
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4. The gradient of 3x^2 is 3(2)x^(2-1) = 6x.
5. The gradient of x^2 +3x +1 is 2x +3.
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If you need to "show work" for these problems individually, use the appropriate values for 'a' and 'n' in the above derivation of the power rule.