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.

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.

To find the linear approximation of 𝑓(𝑥) at 𝑥=3, we'll use the formula for linear approximation:

𝑓(𝑥) ≈ 𝑓(𝑎) + 𝑓′(𝑎) * (𝑥 − 𝑎)

where 𝑓(𝑎) is the value of the function at the point 𝑥=𝑎, and 𝑓′(𝑎) is the derivative of the function evaluated at 𝑥=𝑎.

a.) Given that 𝑓(𝑎) = 5 and 𝑓′(𝑥) = 𝑥/((𝑥^3)+3), we can use 𝑎 = 3:

𝑓(𝑥) ≈ 𝑓(3) + 𝑓′(3) * (𝑥 − 3)

Substituting 𝑓(3) = 5 and 𝑓′(3):

𝑓(𝑥) ≈ 5 + (3 / ((3^3)+3)) * (𝑥 − 3)

Simplifying further:

𝑓(𝑥) ≈ 5 + (3 / 30) * (𝑥 − 3)
≈ 5 + (1 / 10) * (𝑥 − 3)
≈ 5 + 0.1 * (𝑥 − 3)
≈ 5 + 0.1𝑥 − 0.3
≈ 0.1𝑥 + 4.7

Therefore, the linear approximation of 𝑓(𝑥) at 𝑥=3 is 𝑓(𝑥) ≈ 0.1𝑥 + 4.7.

b.) To estimate the value of 𝑓(2.9) using the linear approximation, we substitute 𝑥=2.9 into the linear approximation equation:

𝑓(2.9) ≈ 0.1(2.9) + 4.7
≈ 0.29 + 4.7
≈ 4.99

Comparing this estimated value to the actual value of 𝑓(2.9), we can determine if it is an overestimation or underestimation. However, since the function 𝑓(𝑥) and its behavior around 2.9 are not provided, we cannot definitively determine if it is an overestimation or underestimation without additional information.