For continuity at x = 3, we need k to equal the limit from both sides, which is 9.

The left and right derivatives at x = 0 do not match, hence f is not differentiable at that point.

By the Intermediate Value Theorem, a continuous function on a closed interval takes every value between f(a) and f(b).

Evaluating f(0) = e^0 + 0^2 = 1 + 0 = 1.

f(x) is a polynomial function, which is differentiable everywhere, including at x = 1.

The function is a polynomial and is differentiable everywhere, hence yes.

This is a polynomial function, which is differentiable everywhere on its domain.

Using the limit definition, f'(0) = lim (h -> 0) [(h^2 sin(1/h) - 0)/h] = 0. Thus, f(x) is differentiable at x = 0.

The function is a polynomial and hence differentiable everywhere, including at x = 1.

Both limits as x approaches 0 from the left and right are equal to 1, hence f(x) is continuous at x = 0.

Both limits as x approaches 0 from the left and right are equal to 0, hence f(x) is continuous at x = 0.

Both limits as x approaches 1 from the left and right are equal to 2, hence f(x) is continuous at x = 1.

The left limit is -1 and the right limit is 1, which are not equal. Therefore, f(x) is not continuous at x = 0.

This is a linear first-order differential equation. The integrating factor is e^(2x). Solving gives y = 2 - Ce^(-2x).

Integrating both sides gives y = x^3 + C.

This is a non-linear differential equation. The solution can be found using substitution methods.

Using the integrating factor method, we find y = Ce^(3x) + 2.

The characteristic equation is r^2 + 4 = 0, giving complex roots. The solution is y = C1 cos(2x) + C2 sin(2x).

The characteristic equation is r^2 - 5r + 6 = 0, which factors to (r - 2)(r - 3) = 0, giving the solution y = C1 e^(2x) + C2 e^(3x).

This is a separable equation. Separating variables and integrating gives y = 1/(C - x).

Separating variables and integrating gives y = 1/(C - x).

Using the integrating factor e^(∫(1/x)dx) = x, we can solve the equation.

Finding f'(x) = 3x^2 - 12x + 9 and solving gives critical points at x = 1 and x = 3.

The slope of the tangent at x = 2 is f'(x) = 2x, so f'(2) = 4. The equation of the tangent line is y - 4 = 4(x - 2), which simplifies to y = 4x - 8.

The derivative f'(x) = 2x. At x = 2, f'(2) = 4. The equation of the tangent line is y - 4 = 4(x - 2), which simplifies to y = 2x - 4.

f(x) = e^x is differentiable everywhere as it is an exponential function.

Finding f'(x) = 1/x + 1. Setting f'(x) = 0 gives x = 1 as the minimum point.

f'(x) = 1/x; f'(1) = 1/1 = 1, hence it is differentiable at x = 1.

f(x) = sqrt(x) is not differentiable at x = 0 because the left-hand derivative does not exist.

f(x) is a polynomial function, which is differentiable everywhere.