Question

Sketch the graph of a function that has neither a local maximum nor a local minimum at a point where \(f^{\prime}(x)=0\)

Short answer

Short Answer: The function \(f(x)=x^3\) has a critical point at \(x=0\) with no local maximum or minimum, as the function is increasing both when \(x<0\) and when \(x>0\). This point is an inflection point where the concave shape of the graph changes. The graph of \(f(x)=x^3\) has a cubic curve passing through the origin.

Step-by-step solution

Find the function's derivative.

To find the critical points, we need to take the derivative of the given function. We're using \(f(x)=x^3\), so to find its derivative, we can use the power rule. The power rule states that if \(f(x)=x^n\), then \(f^{\prime}(x)=nx^{n-1}\). $$f^{\prime}(x) = 3x^2$$

Find the critical points.

Critical points are points where the first derivative is zero or undefined. In this case, the derivative is defined for all \(x\). To find where the first derivative is zero, we can set \(f^{\prime}(x)\) to zero and solve: $$3x^2 = 0$$ This equation simplifies to \(x=0\). So, there is a critical point at \(x=0\).

Determine the nature of the critical point.

To determine if the critical point is a local maximum, minimum, or neither, we can use the second derivative test. We need to first find the second derivative, \(f^{\prime\prime}(x)\), and then evaluate it at \(x=0\). To find the second derivative, we can take the derivative of our first derivative: $$f^{\prime}(x) = 3x^2$$ $$f^{\prime\prime}(x) = 6x$$ Now, we can evaluate the second derivative at \(x=0\): $$f^{\prime\prime}(0) = 6(0) = 0$$ Since the second derivative equals \(0\), we cannot use the second derivative test to determine whether the critical point is a local maximum, minimum or neither. We need to analyze the function's behavior around the critical point.

Analyze the behavior of the function around the critical point.

Let's check the slope of the tangent in the intervals \(x<0\) and \(x>0\): For \(x<0\), \(f^{\prime}(x) = 3x^2 > 0\) (tangent is positive, the function is increasing). For \(x>0\), \(f^{\prime}(x) = 3x^2 > 0\) (tangent is positive, the function is increasing). The slopes are both positive in \(x<0\) and \(x>0\), meaning there is neither a local maximum nor minimum at \(x=0\). This point must be an inflection point.

Graph the function.

Now that we have analyzed the behavior of the function, we can sketch the graph of \(f(x)=x^3\). The function has an inflection point at \(x=0\), and we know that the function is increasing in both \(x<0\) and \(x>0\). The graph of the function will look like a cubic curve passing through the origin, with the concave shape changing at the inflection point.