at (3,5), the tangent line has slope f'(3) = 3/30 = .1
so, y-5 = .1(x-3) is the linear approximation near (3,5)
If f(x) is concave up at x=3, the linear approximation will be low.
f''(x) = (3-2x^3)/((x^3+3)^2
f''(3) = (3-18)/100 = -.15
so, f is concave down at x=3, and the tangent line lies above the curve, making it an over-approximation.
a.) Given that f(3)=5 and f'(x)=x/((x^3)+3), find the linear approximation of f(x) at x=3.
b.)If the linear approximation is used to estimate the value of f(2.9), will it be an overestimation or underestimation? Show justification.
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