Question

Find the two \(x\) -intercepts of the function \(f\) and show that \(f^{\prime}(x)=0\) at some point between the two \(x\) -intercepts. $$ f(x)=x(x-3) $$

Short answer

The x-intercepts are at \(x = 0\) and \(x = 3\). \(f'(x) = 0\) for \(x = 1.5\), which is indeed located between the x-intercepts.

Step-by-step solution

Find the x-intercepts

Set the function \(f(x) = x(x - 3)\) equal to 0 and solve for x: \n\n\(f(x) = 0\)\n\(x(x - 3) = 0\)\nThis equation will be true if either \(x = 0\) or \(x = 3\). So the x-intercepts are 0 and 3.

Find the derivative

The derivative of the function \(f(x) = x(x - 3)\) is given by applying the product rule for differentiation: \n\n\(f^{\prime}(x) = 1*(x-3) + x*1\)\nThis simplifies to: \n\n\(f^{\prime}(x) = x - 3 + x = 2x - 3\)

Set the derivative equal to 0 and solve for x

To find where the derivative equals zero, set \(f^{\prime}(x) = 0\) and solve for \(x\): \n\n\(2x - 3 = 0\)\n\(2x = 3\)\n\(x = 3/2\) or \(x = 1.5\)\nSince 1.5 is between the x-intercepts (0 and 3), there's indeed a point on the graph of the function where the slope of the tangent is 0.