Problem 35 - Calculus an Applied Approach

Question

find the second derivative and solve the equation \(f^{\prime \prime}(x)=0\) $$ f(x)=x^{3}-9 x^{2}+27 x-27 $$

Step-by-Step Solution

Answer

The second derivative of the function is \(f''(x) = 6x-18\) and the solution to the equation \(f''(x) = 0\) is \(x = 3\).

Step 1: Compute the First Derivative

To begin, let's calculate the first derivative of the given function, a.k.a \(f'(x)\), by applying the power rule. The power rule states that the derivative of \(x^n\) with respect to x is \(nx^{n-1}\). \n\n So, \[f'(x) = \frac{d}{dx}(x^{3}-9x^{2}+27x-27) = 3x^{2}-18x+27.\]

Step 2: Compute the Second Derivative

Now we calculate the second derivative, which is the derivative of \(f'(x)\). So, \[f''(x) = \frac{d}{dx}(3x^{2}-18x+27) = 6x-18.\]

Step 3: Solve the Equation for \(f''(x) = 0\)

Final step is to solve the equation \(f''(x) = 0\) to find the value of x in order to find the point of inflection. \n\n So, setting \(f''(x) = 0\), we get: \[6x - 18 = 0 \Rightarrow x=3.\]

Key Concepts

Inflection Point

An inflection point on a curve is a point where the curve changes its concavity. This means it transitions from being concave up (shaped like a cup) to concave down (shaped like a cap), or vice versa. At this point, the second derivative of a function, denoted as \(f''(x)\), equals zero. Observing the nature of \(f''(x)\) near this point helps confirm whether a true inflection point exists.

For the given function, we calculated the second derivative as \(f''(x) = 6x - 18\). When we solved the equation \(f''(x) = 0\), the solution gave us \(x = 3\). Here, if \(f''(x)\) changes its sign around \(x = 3\), the function \(f(x)\) has an inflection point at this value accurately indicating a change in concavity. Thus, checking further with sign charts or similar can confirm a true inflection point.

Power Rule

The power rule is an essential technique in calculus for differentiating polynomial functions. It is very straightforward: if you have a term \(x^n\), its derivative is \(nx^{n-1}\). This rule simplifies finding derivatives significantly and enables quick computation even for complex polynomials.

In our original exercise, the power rule was applied to each term in the polynomial function \(f(x) = x^3 - 9x^2 + 27x - 27\). This led to the first derivative:

• For \(x^3\), the derivative is \(3x^2\).
• For \(-9x^2\), the derivative is \(-18x\).
• The linear term \(27x\) becomes \(27\).
• The constant \(-27\) has a derivative of zero.

Thus, combining all these, we obtained \(f'(x) = 3x^2 - 18x + 27\). Using the power rule repeatedly ensures efficiency in differentiating polynomial expressions.

Derivative of a Polynomial Function

Differentiating polynomial functions involves finding the slopes of tangent lines to the function's graph. This is expressed as the first derivative \(f'(x)\), and it provides the rate of change of the function with respect to its variable.

The original polynomial function, \(f(x) = x^3 - 9x^2 + 27x - 27\), undergoes differentiation to produce \(f'(x) = 3x^2 - 18x + 27\). By employing the power rule, each component of the polynomial was handled separately, ensuring accurate calculation of the first derivative.

Moving to the next level with the second derivative, \(f''(x)\), sheds light on the concavity of the graph. Here, \(f''(x) = 6x - 18\) offers insights not just on where the function levels off but also where inflection points might appear—indicative of changes in concavity. Understanding derivatives in this manner is crucial in analyzing the behavior of polynomial functions.