(B) Since y′ = e−x(1 − x) and e−x > 0 for all x, y′ = 0 when x = 1.(D) The slope y′ = 5x^4 + 3 x^2 − 2. Let g = y′. Since g′(x) = 20x^3 + 6 x =
2 x(10x^2 + 3), g′(x) = 0 only if x = 0. Since g′′(x) = 60x^2 + 6, g′′ is always
positive, assuring that x = 0 yields the minimum slope. Find y′ when x =
0.(C) Since 2x − 2yy′ = 0, y′ = . At (4, 2), y′ = 2. The equation of the
tangent at (4, 2) is y − 2 = 2(x − 4).(D) Since , the tangent is vertical for x = 2y. Substitute in the
given equation and solve for y.(D) Since , therefore, dV = 4πr^2 dr. The approximate increase in
volume is dV ≈ 4 π(3^2 )(0.1) in^3.(C) Differentiating implicitly yields 4x − 3y^2 y′ = 0. So . The linear
approximation for the true value of y when x changes from 3 to 3.04 isyat (^) x (^) = 3 + y′at point (3,2) · (3.04 − 3).
Since it is given that, when x = 3, y = 2, the approximate value of y is
or