1/(1+2x)⁵=(1+2x)⁻⁵=1-10x+(-5)(-6)(2x)²/2-... (binomial expansion)=
1-10x+60x²-...
If x is sufficiently small, this approximates to 1-10x.
When x=a=0, both expressions evaluate to exactly 1, so we can expect that values of x close to zero will confirm the linear approximation.
I’m illustrating two different methods here to emphasise the benefits of using linear approximation in questions such as this one.
Method 1
Consider |1/(1+2x)⁵-(1-10x)|<0.05, which determines what values of x will affect the second decimal place so that rounding to one decimal place gives the same result.
Then 60x²<0.05, x²<1/1200 and |x|<0.029 approximately, giving an approximate interval of x in (-0.029,0.029).
Method 2
However, if we use calculus on the linear approximation we get:
y=1-10x, so dy=-10dx, and when dy=0.05, dx=-0.005. In other words, the range for x is [-0.005,0.005].
CHECK
Comparing both methods we can see that the calculus method gives a more reliable range, because, for example, when x=0.029 from the first method 1/(1+2x)⁵=0.754 and 1-10x=0.71 so rounding gives a discrepancy. x=-0.029 gives us 1/(1+2x)⁵=1.348 and 1.290 respectively which both round to 1.3. When x=±0.005 the rounding is accurate, giving 1.051 and 1.05, and 0.951 and 0.95.
