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Calculus
Tutor: None Selected Time limit: 1 Day

Jul 2nd, 2015

Hello!

g(x) has roots at -5, 0 and 4. Be a cubic polynomial, it has a form a*(x+5)*x*(x-4).

36 = g(1) = a*6*1*(-3), so a = -2 and g(x) = -2*(x+5)*x*(x-4).

Now find g'(x):

g'(x) = (-2)*[x(x-4) + (x+5)(x-4) + (x+5)x],
g'(1) = (-2)*[-3 + 6*(-3) + 6] = (-2)*(-15) = 30.

Then the equation of the tangent line is y = g(1) + (x-1)*g'(1), or
y = 36 + (x-1)*30, or y = 30x + 6.

Now approximate g(1.05):
g(1.05) = g(1 + 0.05) ≈ g(1) + 0.05*g'(1) = 36 + 0.05*30 = 36 + 1.5 = 37.5

Please ask if anything is unclear.
Jul 2nd, 2015

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