let fbe a function that is even and continuous on the closed interval [0,3]. the function f and itsderivatives have the properties indicated in the table below. sketch the graph.
x 0 0<x<1 1 1<x<2 2 2<x<3
f(x) 1 (+) 0 (-) -1 (-)
f’(x) DNE (-) 0 (-) DNE (-)
f’’(x) DNE (+) 0 (-) DNE (-)
To sketch the graph of the function, let's analyze the given table and plot the points accordingly.
We know that the function f(x) is even and continuous on the closed interval [0, 3].
First, let's plot the points where the function has specific values:
At x = 0, f(0) = 1. So, we plot the point (0, 1).
Between 0 and 1 (exclusive), f(x) is positive, so we can draw the graph as an increasing line segment. However, we don't know the specific value between 0 and 1, so we leave it unnamed.
At x = 1, f(1) = 0. Thus, we plot the point (1, 0).
Between 1 and 2 (exclusive), f(x) is negative, so we can draw the graph as a decreasing line segment. Again, we leave the specific value between 1 and 2 unspecified.
At x = 2, f(2) = -1. Hence, we plot the point (2, -1).
Finally, from 2 to 3 (inclusive), f(x) is negative, thus drawing a horizontal line segment.
Now, let's analyze the derivatives:
The first derivative, f'(x), is undefined (DNE) at x = 0 and x = 2. So, we mark these points on our graph.
Between 0 and 1 (exclusive), f'(x) is negative. Thus, we draw a decreasing slope leading up to the point (0, 1).
At x = 1, f'(1) = 0. So, we mark this point on our graph as well.
Between 1 and 2 (exclusive), f'(x) is negative. We continue the decreasing slope leading up to (1, 0), and then maintain it until x = 2.
The second derivative, f''(x), is undefined (DNE) at the same points as f'(x), which are x = 0 and x = 2. We indicate these on our graph.
Between 0 and 1 (exclusive), f''(x) is positive. Thus, we draw a concave-upward portion between (0, 1) and (1, 0).
At x = 1, f''(1) = 0. We mark this point on our graph.
Between 1 and 2 (exclusive), f''(x) is negative. We maintain the concave-downward shape from (1, 0) to (2, -1).
Finally, from 2 to 3 (inclusive), f''(x) is negative but undefined at x = 2. We continue the concave-downward shape until (2, -1), and then draw a horizontal line from (2, -1) to (3, -1).
Plotting these points and connecting them with appropriate curves and segments will give you the graph of the given function.