What is the error due to using linear interpolation to estimate the value of sinxsin⁡x at x = \pi/3? your answer should have at least three significant figures, accurate to within 0.1%. (e. g., 1.23 and 3.33e-8 both have three significant

The actual value of sin(π/3) is (√3)/2 ≈ 0.86602540.

If the sine function is approximated by y=x (no error at x = 0), then the error at x=π/3 is ...

... x -sin(x) @ x=π/3

... π/3 -(√3)/2 ≈ 0.18117215 ≈ 0.1812

You know right away this is a bad approximation, because the approximate value is π/3 ≈ 1.04719755, a value greater than 1. The range of the sine function is [-1, 1] so there will be no values greater than 1.

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If the sine function is approximated by y=(x+1-π/4)/√2 (no error at x=π/4), then the error at x=π/3 is ...

... (x+1-π/4)/√2 -sin(x) @ x=π/3

... (π/12 +1)/√2 -(√3)/2 ≈ 0.026201500 ≈ 0.02620

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: