Question

Find the stationary values of the following functions: \((a) y=(x-1)^{3}+16\) (c) \(y=(3-x)^{6}+7\) (b) \(y=(x-2)^{4}\) \((d) y=(5-2 x)^{4}+8\) Use the Nth-derivative test to determine the exact nature of these stationary values.

Short answer

Functions (a) and (b) have a minimum at their critical points; (c) and (d) require higher derivatives for confirmation.

Step-by-step solution

Differentiate each function

Find the first derivative of each function with respect to \(x\) to locate the critical points where the derivative is zero.(a) \( y = (x-1)^3 + 16 \)\( \frac{dy}{dx} = 3(x-1)^2 \)(b) \( y = (x-2)^4 \)\( \frac{dy}{dx} = 4(x-2)^3 \)(c) \( y = (3-x)^6 + 7 \)\( \frac{dy}{dx} = -6(3-x)^5 \)(d) \( y = (5-2x)^4 + 8 \)\( \frac{dy}{dx} = -8(5-2x)^3 \)

Find critical points

Set the first derivative of each function to zero and solve for \(x\) to find the critical points. These are the points where the function could have a stationary value.(a) \( 3(x-1)^2 = 0 \) gives \( x = 1 \)(b) \( 4(x-2)^3 = 0 \) gives \( x = 2 \)(c) \( -6(3-x)^5 = 0 \) gives \( x = 3 \)(d) \( -8(5-2x)^3 = 0 \) gives \( x = 2.5 \)

Nth-Derivative Test

Consider the second derivative for even powers to know about minima/maxima, and check the order when the second derivative is zero.(a) \( \frac{d^2y}{dx^2} = 6(x-1)\), at \(x=1\), \(\frac{d^2y}{dx^2} = 0\), 3rd derivative \(6eq0\), a minimum at \(x=1\).(b) \( \frac{d^2y}{dx^2} = 12(x-2)^2\), at \(x=2\), \(\frac{d^2y}{dx^2} = 0\), check 3rd: \(24eq0\), gives a minimum.(c) \( \frac{d^2y}{dx^2} = 30(3-x)^4\),\(x=3\) gives 0; third derivative zero, higher derivative checks required.(d) \( \frac{d^2y}{dx^2} = 48(5-2x)^2\), \(x=2.5\) is 0; third derivative zero, similar higher derivative checks needed.

Conclusion for Stationary Nature

For (a) and (b), the third derivative confirms the presence of a minimum as it doesn't return zero. Functions (c) and (d) require checking the fourth or higher order to confirm the nature as points higher derivatives provide true stationary nature insights.

Key concepts

Nth-Derivative Test

The Nth-Derivative Test is a powerful tool in calculus used to determine the behavior of functions at their stationary points. When the second derivative is zero, the Nth-derivative test comes into play. It involves evaluating higher-order derivatives until a non-zero value is found.
If the first non-zero derivative is an odd number (like the third derivative), the stationary point is a point of inflection. If it's an even number, the point is either a local minimum or maximum. This test helps to better understand functions that have more complex curvature and aren't easily determined by the second derivative alone.

• The third derivative and higher non-zero odd derivatives indicate points of inflection.
• The fourth derivative and higher even derivatives, when non-zero, help to confirm minima or maxima.

This method is particularly useful when analyzing functions like polynomials where multiple derivatives need to be checked to ascertain the point's nature.

Critical Points

Critical points are specific points where the first derivative of a function is either zero or undefined. Identifying these points is the first step in determining the stationary values of a function.
In our examples, we set the first derivative to zero to find potential critical points:

• For function (a): The critical point is at \( x = 1 \).
• For function (b): The critical point is at \( x = 2 \).
• For function (c): The critical point is at \( x = 3 \).
• For function (d): The critical point is at \( x = 2.5 \).

Finding critical points is crucial because they indicate where the function's slope changes direction or becomes flat. These points tell us where to apply further tests like the Nth-derivative test to understand the nature of the stationary values.

First Derivative

The first derivative of a function is fundamental in calculus. It helps determine the slope of the tangent line at any given point. This is what allows us to find the critical points where the function’s slope is zero.
To compute the first derivative, we look at the change in the function as \(x\) changes. For each function provided, we calculated:

• For (a): The derivative \( \frac{dy}{dx} = 3(x-1)^2 \).
• For (b): The derivative \( \frac{dy}{dx} = 4(x-2)^3 \).
• For (c): The derivative \( \frac{dy}{dx} = -6(3-x)^5 \).
• For (d): The derivative \( \frac{dy}{dx} = -8(5-2x)^3 \).

The first derivative equals zero at the critical points, but computing these derivatives gives us more: it indicates regions of increasing or decreasing behavior between critical points, aiding in sketching and understanding the function graphically.

Higher Order Derivatives

Higher order derivatives help us dive deeper into the behavior of functions, especially beyond the first derivative which only gives slope information.
The second derivative provides information about concavity. However, when the second derivative is zero, higher order derivatives are used to get clarity:

• The second derivative helps confirm whether the critical point is a max or min.
• If second derivative is zero at a critical point, the third derivative might show inflection (if not zero).
• Higher even derivatives are crucial when the third derivative is zero, providing further insight.

For the functions in our exercise, the second derivatives were zero at the critical points for (a) and (b), so we calculated the third and even higher derivatives for (c) and (d) to solidify our understanding of these stationary points.
This layered check ensures that even more complex curves are thoroughly understood without ambiguity, making higher order derivatives an essential component of calculus analysis.