comparing secant slope through A:(xA,yA)A:(xA,yA) and B:(xB,yB)B:(xB,yB)
taking AA and BB as points “on” the function y=f(x)y=f(x), we have yA=f(xA)yA=f(xA) and yB=f(xB)yB=f(xB)
so A:(xA,f(xA))A:(xA,f(xA)) and B:(xB,f(xB))B:(xB,f(xB))
secant slope: ΔyΔx=f(xB)−f(xA)xB−xAΔyΔx=f(xB)−f(xA)xB−xA
instantaneous (or “tangent”) slope requires AA and BB to be “infinitely close”
fix the point where we want instantaneous slope: A:(a,f(a))A:(a,f(a))
let point BB vary, so that it can “approach” AA: B:(x,f(x))B:(x,f(x))
then secant slope: ΔyΔx=f(x)−f(a)x−aΔyΔx=f(x)−f(a)x−a
note that ΔyΔxΔyΔx is undefined if we let x=ax=a
in other words, we cannot make BB and AA be the same point... of course, slope needs two points!
Compute secant slopes of f(x)f(x) through AA
approximate the instantaneous slope at: A:(−3,27−−√)A:(−3,27)
secant slope from the left: secant slope from the right:
xx f(x)f(x) ΔyΔxΔyΔx xx f(x)f(x) ΔyΔxΔyΔx
Your answers must be given as decimal approximations.
When completing the table, make sure your answers are accurate to at least 4 decimal places.
Be careful when computing the slope, you may need more than 4 decimal places for the yy-coordinate in order for the slope to achieve the required accuracy.
Estimate the instantaneous slope of f(x)f(x) at AA
Usi1 AnswerHomework Help3 weeks ago
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