Asked by Joe
Use linear approximation, i.e. the tangent line, to approximate (1/0.504) as follows: Find the equation of the tangent line to f(x)=1/x at a "nice" point near 0.504. Then use this to approximate (1/0.504).
Answers
Answered by
MathMate
The idea behind the approximation is:
y(x)=1/x
y'(x)=-1/x²
Let x=x0+Δx
y(x) = y(x0)+y'(x0)*Δx
=y(x)-Δx / (x0 ²)
Now apply x0=0.5, Δx=0.004
y(0.504)=y(0.5)-0.004/0.5²
=2-0.016
=1.984
compare to approximate actual value of
1.984126984126984
y(x)=1/x
y'(x)=-1/x²
Let x=x0+Δx
y(x) = y(x0)+y'(x0)*Δx
=y(x)-Δx / (x0 ²)
Now apply x0=0.5, Δx=0.004
y(0.504)=y(0.5)-0.004/0.5²
=2-0.016
=1.984
compare to approximate actual value of
1.984126984126984
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