Question

In Exercises \(9-20\), find the points of inflection and discuss the concavity of the graph of the function. \(f(x)=x^{3}-6 x^{2}+12 x\)

Short answer

The point of inflection of the function \(f(x)=x^{3}-6 x^{2}+12 x\) is \((2,8)\). The graph is concave down when \(x<2\) and concave up when \(x>2\).

Step-by-step solution

Differentiate the function

Find the first and the second derivatives of the function.The first derivative \(f'(x)\) is obtained by differentiating the function \(f(x)\) term by term. Differentiate \(x^{3}\), \(-6 x^{2}\), and \(12x\) to obtain \(f'(x)=3x^{2}-12x+12\).Then differentiate \(f'(x)\) to obtain the second derivative \(f''(x)\). Differentiate \(3x^{2}\), \(-12x\), and \(12\) to obtain \(f''(x)=6x-12\).

Find potential points of inflection

To find the points of inflection, we set the second derivative \(f''(x)=6x-12\) equal to zero and solve for \(x\). This results in \(x=2\). Therefore, \(x = 2\) is a potential point of inflection.

Verify the points of inflection

To verify that \(x = 2\) is a point of inflection, we need to check that \(f''(x)\) changes sign at this point. Choose a test point to the left of \(x = 2\), say \(x = 1\), and to the right, say \(x = 3\). Evaluate \(f''(x)\) at these test points.If \(x=1\), \(f''(x) = 6(1)-12 = -6 < 0\) - the function is concave down.If \(x=3\), \(f''(x) = 6(3)-12 = 6 > 0\) - the function is concave up.Since the graph changes from concave down to concave up, \(x = 2\) is indeed a point of inflection.

Find the y-coordinate of the point of inflection

To find the y-coordinate of the point of inflection, substitute \(x = 2\) into the original function \(f(x)\). This yields \(f(2) = 2^3 - 6*2^2 + 12*2 = 8 - 24 + 24 = 8\). Therefore, the point of inflection is \((2, 8)\).