A-Use a linear approximation to approximate 3.0016 as follows: The linearization

Question

A-Use a linear approximation to approximate 3.0016 as follows:

The linearization L(x)  to f(x)=x^6 at a=3 can be written in the form L(x)=mx+b

where m is:  and where b is: ............
Using this, the approximation for 3.0016 is ...................

B-Let y=2tanx.

(a) Find the differential   dy=  dx. ........

(b) Evaluate  dy and ?y when x=?/4 and dx=?0.2 ..........

dy=  and ?y= ..............

C-Let y=4x?.

Find the differential dy=  dx. .............

Find the change in y, ?y when x=5 and ?x=0.2 ........

Find the differential dy when x=5 and dx=0.2.........

Explanation / Answer

A) Given function f(x)=x^6    at a=3.

Linear approximation is:

L(x)=f(a)+f1(a)(x-a)

=x6+6.x5(x-3)

=36+6(35)(x-3)

=729+6.243(x-3)

=729+1458(x-3)

=729+1458x-438-4374

=145, b=-8x-3645   it like a mx+b

so m=1452,   b=-3645.

Approximation value for 3.0016.

L(3.0016)=1458(3.0016)-3645

                =4376.33-3645

                 =731.33

B) Given function: y=2tanx

    dy=2 Sec2x.dx

a) dy=dx.2 Sec2 x

c) Let y=4x

    dy=4dx

      dy=dx.4

Find the change in y=y-? and x=5, ?x=0.2

   y-?y=4.(x-?)

y-?Y=4.(5-0.2)

y-?y=4(4.8)

    =19.2

Differential dy when x=5, dx=0.2

dy=4dx

      =4(0.2)    =0.8

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