Finding Derivatives at a Point

If the tangent line to f(x) at x = 2 has a slope of -5, state the derivative f'(2) and whether the graph is rising or falling at that point.

f'(2) = -5. The graph is falling because the derivative is negative.

Explain why the derivative of any constant function f(x) = c is always 0.

A constant function produces a horizontal line with no vertical change. The change in y is always 0, so the slope is 0 everywhere.

Set up the difference quotient for f(x) = 5x at the point x = 4 using a step of h.

[5(4 + h) - 5(4)] / h, which simplifies to [20 + 5h - 20] / h.

Simplify the expression [5(4 + h) - 5(4)] / h and evaluate the limit as h approaches 0.

5h / h cancels to 5. The limit as h approaches 0 is 5.

Using the limit definition, compute the derivative of f(x) = x^2 at x = -2.

Set up [(-2 + h)^2 - (-2)^2] / h = [4 - 4h + h^2 - 4] / h = (-4h + h^2) / h = -4 + h. Letting h go to zero yields -4.

A car's position in meters is given by s(t) = 3t^2. Find the instantaneous velocity at t = 2 seconds using limits.

Set up [3(2 + h)^2 - 3(2)^2] / h = [3(4 + 4h + h^2) - 12] / h = (12h + 3h^2) / h = 12 + 3h. As h approaches 0, velocity is 12 m/s.

At a certain point on a curve, the derivative is zero. Describe the appearance of the tangent line at that point.

The tangent line is completely horizontal with a slope of zero.