Question

Suppose h(3) = 4 and h'(3) = -2 Use a tangent line approximation to estimate h(2.85).

Answer 1

if \(x\approx x_0\) then \(f(x)\approx f(x_0)+f'(x_0)\cdot(x-x_0)\) -- this is a first-order approximation

Answer 2

The idea here is to use the linearization \(L(x) = h^{\prime}(3)(x-3)+h(3)\) to approximate \(h(x)\) at \(x=3\). So \(h(x)\approx L(x) = -2(x-3)+4 = -2x+10\). Hence, \(h(2.85) \approx L(2.85)=\ldots\)

Can you take things from here?

Answer 3

I mean near \(x=3\), not at \(x=3\).

Answer 4

so: $$h(2.85)\approx h(3)+h'(3)\cdot(2.85-3)=4-2(-0.15)=4+0.3=4.3$$