Δy = f(1.1)-f(1) = 0.4641
Δy ≈ y' Δx = 4x^3*0.1 = 0.41
you are low because the graph is concave up at x=1. The y' approximation follows the tangent line at x=1, rather than going exactly along the curve.
Use the information to evaluate and compare Δy and dy. (Round your answers to three decimal places.)
y = x4
x = 1
Δx = dx = 0.1
I'm totally lost
2 answers
y=x^4
dy=4x^3 dx=4(1)(.1)=.4
deltay= (x+dx)^4-x^4
= x^4+4dx*x^3+6dx^2*x^2+4dx^3*x+ dx^4 -x^4
check that
= 4(.1)1+6(.01)1 + 4(.001)1+(.0001)=
=.4641 check
dy=4x^3 dx=4(1)(.1)=.4
deltay= (x+dx)^4-x^4
= x^4+4dx*x^3+6dx^2*x^2+4dx^3*x+ dx^4 -x^4
check that
= 4(.1)1+6(.01)1 + 4(.001)1+(.0001)=
=.4641 check
1 answer