Verify the given linear approximation at a = 0. Then determine the values of x for which the linear approximation is accurate to within 0.1. (Enter the answer using interval notation. Round to 3 decimal places)
fourth root of 1 + 2x = 1 + (1/2)x
fourth root of 1 + 2x = 1 + (1/2)x
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f(x) = ∜(1+2x)
f(0) = ∜1 = 1
f'(x) = 1/4 (1+2x)^(-3/4) * 2
f'(x) = 1/(2∜(1+2x)³)
f'(0) = 1/2
Linear approximation:
y = 1 + 1/2 x
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Now we need to determine values of x for which the linear approximation is accurate to within 0.1
First, let's graph the original function and its linear approximation:
http://www.wolframalpha.com/input/?i=y+%…
As you can see, the vertical distance between the curve and the line indicate the accuracy of the approximation (the closer they are, the more accurate the approximation). Now let's also graph a line parallel to our linear approximation, but 0.1 units below it vertically
y = 1 + 1/2 x − 1/10 = 9/10 + 1/2 x
http://www.wolframalpha.com/input/?i=y+%…
The two points where this new line intersects the graph of our original function will be points where the approximation is off by 0.1. Any points in between will be accurate to within 0.1. So we just need to find where this line intersects graph of original function, i.e. solve:
∜(1+2x) = 9/10 + 1/2 x
There is no easy way to do this. You can use an approximation method such as Newton's method, or you could use a graphing calculator.
http://www.wolframalpha.com/input/?i=%E2…
Using graphing calculator, we get x = −0.368935, x = 0.677669
So linear approximation will be accurate to within 0.1
when x is on interval (−0.369, 0.678)
f(0) = ∜1 = 1
f'(x) = 1/4 (1+2x)^(-3/4) * 2
f'(x) = 1/(2∜(1+2x)³)
f'(0) = 1/2
Linear approximation:
y = 1 + 1/2 x
---------------------------------
Now we need to determine values of x for which the linear approximation is accurate to within 0.1
First, let's graph the original function and its linear approximation:
http://www.wolframalpha.com/input/?i=y+%…
As you can see, the vertical distance between the curve and the line indicate the accuracy of the approximation (the closer they are, the more accurate the approximation). Now let's also graph a line parallel to our linear approximation, but 0.1 units below it vertically
y = 1 + 1/2 x − 1/10 = 9/10 + 1/2 x
http://www.wolframalpha.com/input/?i=y+%…
The two points where this new line intersects the graph of our original function will be points where the approximation is off by 0.1. Any points in between will be accurate to within 0.1. So we just need to find where this line intersects graph of original function, i.e. solve:
∜(1+2x) = 9/10 + 1/2 x
There is no easy way to do this. You can use an approximation method such as Newton's method, or you could use a graphing calculator.
http://www.wolframalpha.com/input/?i=%E2…
Using graphing calculator, we get x = −0.368935, x = 0.677669
So linear approximation will be accurate to within 0.1
when x is on interval (−0.369, 0.678)
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OK. If you take the quantity (1+2x)^(1/4) = 1 + x/2 and take both sides to the 4th power you get
1+2x = ((2+x)/2)^4. If x=0 then the equation holds. You can also graph the difference of the two sides to get the same answer.
1+2x = ((2+x)/2)^4. If x=0 then the equation holds. You can also graph the difference of the two sides to get the same answer.