Question

Sketch the graph of a function \(f\) having the given characteristics. \(f(0)=f(2)=0\) \(f(3)\) is defined. $$ f^{\prime}(x)>0 \text { if } x<1 $$ $$ \begin{array}{ll} f^{\prime}(x)<0 \text { if } x<3 & f^{\prime}(1)=0 \\ f^{\prime}(3) \text { does not exist. } & f^{\prime}(x)<0 \text { if } x>1 \end{array} $$ $$ \begin{array}{ll} f^{\prime}(x)>0 \text { if } x>3 & f^{\prime \prime}(x)<0 \\ f^{\prime \prime}(x)<0, x \neq 3 & \end{array} $$ 51\. \(f(2)=f(4)=0\) \(f(3)\) is defined. \(f^{\prime}(x)<0\) if \(x<3\) \(f^{\prime}(3)\) does not exist. \(f^{\prime}(x)>0\) if \(x>3\) \(f^{\prime \prime}(x)<0, x \neq 3\) 52\. \(f(0)=f(2)=0\) \(f^{\prime}(x)>0\) if \(x<1\) \(f^{\prime}(1)=0\) \(f^{\prime}(x)<0\) if \(x>1\) \(f^{\prime \prime}(x)<0\)

Short answer

The function graph starts from the origin (0,0), increases in value until around x=1(from first derivative test), then decreases with a sharp turn at x=3(from the rule that \(f'(3)\) does not exist). It continues to decrease until it hits (2,0). After that, it increases again when x>3. The graph has a mostly concave down shape(from the second derivative being less than 0).

Step-by-step solution

Plot given function value points

Start by drawing the x-y axes and mark the given points \(f(0) = 0\) and \(f(2) = 0\) on the graph.

Analyze and mark regions of increase and decrease

The derivative\(f'\) gives us information about where the function is increasing and where it is decreasing. Mark the regions where \(f'(x) > 0 \) (x < 1 and x > 3), indicating that the function is increasing, and where \(f'(x) < 0\) (1 < x < 3 and x > 1), indicating that the function is decreasing.

Identify points where the derivative doesn't exist

From the given conditions, we know that \(f'(3)\) doesn't exist. This suggests a possible point of inflection or discontinuity in the function at x = 3. Since \(f(3)\) is defined, a sharp turn in the function could be expected at this point.

Determine concavity from second derivative

The second derivative \(f''\), tells us about the concavity of the function. In this case, \(f''(x) < 0\) for all \(x ≠ 3\). This means that the function is concave down everywhere except possibly at x=3, adding another layer to our sketch.

Sketch the function

Using all these insights, sketch the graph that meets all the criteria. The function should start at (0,0), increase until it reaches its peak around x=1, then decrease with a sharp turn at x=3. The entire sketch should display a concave down shape due to the second derivative conditions.

Key concepts

First Derivative Test

The first derivative test is a powerful tool to determine whether a function's critical points are relative minimums, maximums, or neither. In essence, it examines the sign changes in the first derivative, which is represented as \( f'(x) \), around the critical points.

For instance, if \( f'(x) \) changes from positive to negative as you pass through a critical point, that critical point is a relative maximum. Conversely, if the first derivative transitions from negative to positive, you've got a relative minimum. If the first derivative does not change signs, then it is neither maxima nor minima, but it could be an inflection point.

In our given problem, the first derivative \( f'(1)=0 \) indicates a critical point at \( x=1 \). Since \( f'(x) > 0 \) when \( x < 1 \) and \( f'(x) < 0 \) when \( x > 1 \), the function increases before this point and decreases after. Therefore, according to the first derivative test, \( x=1 \) is a relative maximum. Similarly, since \( f'(x) \) is positive before and after \( x=3 \), where the derivative does not exist, we suspect a corner or cusp at this point rather than a relative extremum.

Concavity and Inflection Points

When discussing the nature of graphs, concavity refers to the direction in which a curve bends. Think of a concave up graph as shaped like a cup that can hold water, while a concave down graph looks like an arch or an upside-down cup. Inflection points mark the transition between these two states of concavity.

To determine concavity, one must look at the sign of the second derivative, \( f''(x) \). If it’s positive, the function is concave up; if negative, concave down. In our exercise, since \( f''(x) < 0 \) except at \( x=3 \), the graph is concave down everywhere else. Importantly, the point where the second derivative is undefined or changes sign could be an inflection point, prompting the curve to switch concavity.

However, since the second derivative remains negative on both sides of \( x=3 \), there's no change in concavity. Although there's no inflection point, the nondifferentiability at \( x=3 \) introduces a sharp turn but maintains the concave down orientation throughout the sketch.

Continuity of Functions

Now let's delve into the notion of continuity. A continuous function, informally speaking, can be drawn without lifting the pencil from the paper. Continuity at a point means the function is defined there, its limit as you approach the point exists, and the limit equals the actual function value at that point.

In the context of the problem, continuity plays a key role. The function values at \( f(0)=0 \) and \( f(2)=0 \) and the fact that \( f(3) \) is defined suggest that there are no jumps or gaps at these points. Still, this does not preclude potential discontinuity elsewhere.

The catch is the derivative \( f'(3) \) not existing while \( f(3) \) is still defined. This scenario is a hallmark of a continuous function displaying a corner or cusp at \( x=3 \). Such a feature breaks the smoothness of the graph but doesn't interrupt the function's continuity; thus, our sketch reflects a continuous curve with a distinctive corner at \( x=3 \).