How do I make a graph of
f'(-1) = f'(1) = 0, f'(x) > 0 on (-1,1),
f'(x) < 0 for x < -1, f'(x) > 0 for x >1
Thanks.
if the derivative is postive on -1 to 1, you have a curve that shope upward as x increases.
If the derivative is zero at -1, and 1, thekn at -1 the curve is a local min, and at 1, the curve is a local max.
For x<-1, the function decreases for increasing x (or as x becomes greater negative, the curve goes upward).
For x>1, the curve again goes upward. A neat point at x=1. As the curve approaches 1 from the left, it levels off, then as x increases, it starts back up again, as a stairstep.
To make a graph of the given conditions, follow these steps:
1. Start by drawing a horizontal x-axis and a vertical y-axis on a piece of graph paper.
2. Plot the point (-1, f(-1)) on the graph. Since f'(-1) = 0, this point represents a local minimum. Depending on the specific function, you may need to determine the y-coordinate of this point using additional information or calculations.
3. Similarly, plot the point (1, f(1)) on the graph. Since f'(1) = 0, this point represents a local maximum. Again, you might need more information to determine the exact y-coordinate.
4. From these two points, draw a smooth curve that increases as x approaches 1 from the left and then starts to decrease after passing 1. The curve should have a steep slope near -1 and become nearly horizontal near 1 in order to represent a local minimum and a local maximum, respectively.
5. On the left side of -1, draw a smooth curve that decreases as x becomes more negative. This represents the condition that f'(x) < 0 for x < -1. The curve should have a downward slope.
6. On the right side of 1, draw another smooth curve that increases as x becomes larger than 1. This represents the condition that f'(x) > 0 for x > 1. The curve should have an upward slope.
Remember that the description above assumes a general shape for the graph based on the given conditions. The specific shape and exact y-values may depend on the function being graphed.