The linear approximation for the expression indicated as per the attached question is given by:
[tex]L\left ( x \right );[/tex]
[tex]L\left ( x \right )=f\left ( a \right )+f{}'\left ( a \right )\left ( x-a \right )[/tex].
What is the justification for the above answer?
[tex]f\left ( x \right )=\sin ^{2}x,\, \, a=2\pi[/tex]
[tex]f\left ( 2\pi \right )=\sin ^{2}2\pi =0[/tex]
[tex]f{}'\left ( x \right )=\frac{\mathrm{d} }{\mathrm{d} x}\left [\sin ^{2}x \right ][/tex]
[tex]=2\sin x\times \frac{\mathrm{d} }{\mathrm{d} x}\left [\sin x \right ][/tex]
[tex]=2\sin x\times \cos x[/tex]
[tex]=\sin 2x[/tex]
[tex]f{}'\left ( 2\pi \right )=\sin \left (2\times 2\pi \right ) =0[/tex]
[tex]L\left ( x \right )=f\left ( 2\pi \right )+f{}'\left ( 2\pi \right )\left ( x-2\pi \right )[/tex]
[tex]=0+0\left ( x-2\pi \right )[/tex]
= 0
Step 2 : [tex]f\left ( x \right )=x+x^{4},\, \, a=0[/tex]
f (0) = - + 0⁴ = 0
[tex]f{}'\left ( x \right )=\frac{\mathrm{d} }{\mathrm{d} x}\left [x+x^{4} \right ][/tex]
[tex]=\frac{\mathrm{d} x}{\mathrm{d} x}+\frac{\mathrm{d} }{\mathrm{d} x}\left [x^{4} \right ][/tex]
= 1 + 4x³
f' (0) = 1 + 4 x 0³
= 1
L (x) = f(0) + f' (0) (x-0)
= 0 + 1 * x
= x
Learn more about Linear Approximation:
https://brainly.com/question/24318076
#SPJ4
