Question

For each of the following functions of one variable, determine all local maxima and minima and indicate points of inflection (where $f^{\prime \prime }=0$ ): a. $f(x)=4x^3-12x$ b. $f(x)=4x-x^2$ c. $f(x)=x^3$

Short answer

Question: Identify the local maxima, local minima, and points of inflection for the following functions: (a) $f(x)=4x^3-12x$, (b) $f(x)=4x-x^2$, and (c) $f(x)=x^3$. Answer: (a) f(x) has a local maximum at x = -1, a local minimum at x = 1, and a point of inflection at x = 0. (b) f(x) has a local maximum at x = 2 and no point of inflection. (c) f(x) has no local maximum or minimum, but has a point of inflection at x = 0.

Step-by-step solution

Find the first derivative f'(x)

Compute the first derivative of the function using the power rule, $f^{\prime }(x)=12x^2-12$.

Apply the first derivative test

Find critical points by setting the first derivative equal to zero and solving for x: $12x^2-12=0$. Factor out 12: $12(x^2-1)=0$. This gives us the critical points: $x=-1$ and $x=1$. Use the first derivative test to analyze these critical points: - For $x<-1$, $f^{\prime }(-2)=12$, so the function is increasing. - For $-1<x<1$, $f^{\prime }(0)=-12$, so the function is decreasing. - For $x>1$, $f^{\prime }(2)=12$, so the function is increasing. Therefore, we have a local maximum at $x=-1$ and a local minimum at $x=1$.

Find the second derivative f''(x)

Compute the second derivative by differentiating f'(x), $f^{″}(x)=24x$.

Determine the points of inflection

Set the second derivative equal to zero: $24x=0$. Solve for x: $x=0$. So, the function has a point of inflection at $x=0$. For function b: $f(x)=4x-x^2$

Find the first derivative f'(x)

Compute the first derivative using the power rule: $f^{\prime }(x)=4-2x$.

Apply the first derivative test

Find critical points by setting the first derivative equal to zero: $4-2x=0$. Solve for x: $x=2$. Analyze the critical point using the first derivative test: - For $x<2$, $f^{\prime }(1)=2$, so the function is increasing. - For $x>2$, $f^{\prime }(3)=-2$, so the function is decreasing. Therefore, we have a local maximum at $x=2$.

Find the second derivative f''(x)

Compute the second derivative by differentiating f'(x), $f^{″}(x)=-2$.

Determine the points of inflection

Since the second derivative is constant, there is no point of inflection. For function c: $f(x)=x^3$

Find the first derivative f'(x)

Compute the first derivative using the power rule: $f^{\prime }(x)=3x^2$.

Apply the first derivative test

Find critical points by setting the first derivative equal to zero: $3x^2=0$. Solve for x: $x=0$. However, since the function decreases for $x<0$ and increases for $x>0$, there is no local maximum or minimum at $x=0$.

Find the second derivative f''(x)

Compute the second derivative by differentiating f'(x), $f^{″}(x)=6x$.

Determine the points of inflection

Set the second derivative equal to zero: $6x=0$. Solve for x: $x=0$. So, the function has a point of inflection at $x=0$.

Key concepts

Local maxima and minima

In the world of calculus, local maxima and minima are pivotal in understanding the behavior of functions. To put it simply, a local maximum is a point where the function reaches its highest value in a certain interval, while a local minimum is where it reaches its lowest. Identifying these points aids in sketching the graph of the function and finding where it increases or decreases.

To determine local maxima and minima, we often rely on the *First Derivative Test*. This involves:

• Computing the first derivative of the function.
• Setting the derivative equal to zero to find the critical points.
• Analyzing these points to see where the function changes from increasing to decreasing, or vice versa.

For instance, in the function given as an example, $f(x)=4x^3-12x$, the maximum is at $x=-1$ and the minimum at $x=1$. These critical points mark local extremes where the function changes direction, providing valuable insights into its shape.

Critical points

Critical points are specific values in the domain of a function where its derivative is zero or undefined. At these points, a function may experience a change in its increasing or decreasing behavior. This makes them particularly important when investigating the function's extrema (like maxima and minima).

To locate critical points:

• First, compute the derivative of the function.
• Set this derivative to zero and solve for $x$.
• The solutions you find are the critical points of the function.

In the example $f(x)=4x^3-12x$, the critical points are found to be $x=-1$ and $x=1$. At $x=-1$, the function changes from increasing to decreasing, indicating a local maximum. At $x=1$, the switch from decreasing to increasing signifies a local minimum. These critical points play an essential role in defining the function's overall behavior.

Inflection points

Inflection points are fascinating aspects of calculus that occur where a function changes its concavity. In simpler terms, it is where a curve switches from being concave upwards (like a cup) to concave downwards (like a cap), or vice versa. Detecting these points helps in understanding how steep a curve gets or how it "bends" on a graph.

To identify inflection points:

• Start by finding the second derivative of the function.
• Set this second derivative equal to zero and solve for $x$.
• The solutions indicate potential inflection points where the concave change might occur.

Taking the previous function $f(x)=4x^3-12x$ as an example, the second derivative $f^{″}(x)=24x$ equals zero at $x=0$. This tells us that $x=0$ is an inflection point, showing us where the curve shifts from being concavely "up" to "down" or vice versa. Understanding these points doesn't just reveal more about the function's geometry but also allows for a deeper comprehension of how functions behave.