Question

let f(x)= 3^x - x^3 find the values of x (correct to thousandths) such that the tangent to the curve is parallel to the secant through (0, 1) and (3, 0)

Answer 1

Since f is continuous on [0,3] and differentiable on (0,3), we can use the Mean Value Theorem. f ' (x) = 3^x * ln3 - 3x^2. This represents the slope of the tangent.
The slope of the secant is (0-1)/3-0) = -1/3. Now use your calculator to solve f ' (x) = -1/3, choosing the value(s) of x only on the given interval [0,3].