f(1.1) ≈ 0.23

f(1.2) ≈ -0.17

There is a change of sign, so a root must exist between the two.

]]>2)

f(x) = x² + sin(x/2) f(0) = 0

f'(x) = 2x + (1/2)cos(x/2) f'(0) = 1/2

f''(x) = 2 - (1/4)sin(x/2) f''(0) = 0

f'''(x) = -(1/8)cos(x/2) f'''(0) = -1/8

f''''(x) = (1/16)sin(x/2) f''''(0) = 0

f'''''(x) = (1/32)cos(x/2) f'''''(0) = 1/32

0 + [(1/2)x]/1! + 0x²/2! - [(1/8)x³]/3! + [0x^(4)]/4! + [(1/32)x^5]/5!

]]>x - f(x)/f '(x)

Oh, and be sure to use radians because degrees will not give a solution.

You can use degrees but then your equation will be: y = 2x - tan(180x/pi)

]]>PLEASE HELP!!!

Show that the equation 2x-tanx=0 has a root in the interval [1.1, 1.2]. hence find this root correct to 2 decimal places.

b) Find the Macluarin series expansion for f(x) = x2 + sin (x/2) up to and including the term in x5.

Please HELP........

Deeply appreciated.

Thank you

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