Δf = f(3.02) - f(3)
f(x) = √21x + 28
f(3) = √21 x 3 + 28
= √63 + 28
= √91
f(3.02) - f(3)
√91.4 - √91 = 9.561 - 9.539 = 0.022.
Δf ≈ 0.............ANS
f(x)=sqr(21x+28)
df/dx=del(f)/del(x)+e
where
del(f)/del(x)->df/dx & e>0->0,
as del(x)->0
Thus,
del(f)=(df/dx)del(x).
Now, at x=3 & del(x)=0.02
=>
df/dx=
21/[2sqr(21*3+28)]=
1.1007
del(f)=1.1007*0.02=0.022
approximately.
f(3.02) - f(3) =
√(21(3.02) + 28) - √(21(3) + 28) =
√91.06 - √91 =
0.00314
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Δf = f(3.02) - f(3)
f(x) = √21x + 28
f(3) = √21 x 3 + 28
= √63 + 28
= √91
f(3.02) - f(3)
√91.4 - √91 = 9.561 - 9.539 = 0.022.
Δf ≈ 0.............ANS
f(x)=sqr(21x+28)
df/dx=del(f)/del(x)+e
where
del(f)/del(x)->df/dx & e>0->0,
as del(x)->0
Thus,
del(f)=(df/dx)del(x).
Now, at x=3 & del(x)=0.02
=>
df/dx=
21/[2sqr(21*3+28)]=
1.1007
=>
del(f)=1.1007*0.02=0.022
approximately.
f(3.02) - f(3) =
√(21(3.02) + 28) - √(21(3) + 28) =
√91.06 - √91 =
0.00314