Question

sketching a derivative in exercises 43 - 48, sketch the graph of f. explain how you found your answer. 43. 44.

Explanation:

Step1: Recall the definition of the derivative

The derivative $f^{\prime}(x)$ represents the slope of the tangent line to the graph of $y = f(x)$ at the point $(x,f(x))$.

Step2: Analyze the graph of $f$ in exercise 43

The graph of $y = f(x)$ in exercise 43 is a straight - line. The equation of a straight - line is of the form $y=mx + b$, where $m$ is the slope. For a non - vertical straight - line, the slope is constant. Using two points on the line, say $(0,1)$ and $(1,0)$, the slope $m=\frac{0 - 1}{1-0}=- 1$. So, $f^{\prime}(x)=-1$ for all $x$. The graph of $y = f^{\prime}(x)$ is a horizontal line at $y = - 1$.

Step3: Analyze the graph of $f$ in exercise 44

The graph of $y = f(x)$ in exercise 44 is a vertical line. The function $y = f(x)$ is not a function in the traditional sense (it fails the vertical line test for a function of $x$). The slope of a vertical line is undefined. The derivative $f^{\prime}(x)$ does not exist for any $x$ value associated with this vertical line. We cannot sketch a graph of $y = f^{\prime}(x)$ in the real - valued function sense since the derivative is not defined for any $x$ in the domain of this "vertical - line relation".

Answer:

For exercise 43: The graph of $y = f^{\prime}(x)$ is a horizontal line at $y=-1$. For exercise 44: The derivative $f^{\prime}(x)$ is not defined, so we cannot sketch a real - valued graph of $y = f^{\prime}(x)$.