3. Discuss the graph of y=x/3+2+3/x. Determine the values of x for which the graph is
a. Continuous
b. Rising
c. Falling
d. Concave upward
e. Concave downward
f. Has a relative maximum
g. Has a relative minimum
If you combine terms you see that it's just
y = (x+3)^2/3x
y' = 1/3 - 3/x^2
y'' = -6/x^3
Continuous except where denominator is zero
rising/falling where y' is positive/negative
concave up where y'' > 0
max/min where y' = 0
Oops
y'' = 6/x^3
To discuss the graph of y = (x/3) + 2 + (3/x), we can analyze its different characteristics.
a. Continuous: The graph of a function is continuous if there are no breaks, holes, or jumps in the graph. In this case, the function y = (x/3) + 2 + (3/x) is continuous for all values of x except x = 0 (since division by zero is undefined).
b. Rising: The graph is considered to be rising when the slope of the curve is positive. Given the equation y = (x/3) + 2 + (3/x), the slope of the curve is positive for all values of x. Therefore, the graph is always rising.
c. Falling: The graph is considered to be falling when the slope of the curve is negative. However, in this case, since the slope of the curve is always positive, the graph does not fall.
d. Concave upward: A graph is concave upward when its second derivative is positive. To find the second derivative of y = (x/3) + 2 + (3/x), we need to differentiate it twice. The second derivative here is 6/x^3. The second derivative is positive when x > 0, so the graph is concave upward for x > 0.
e. Concave downward: A graph is concave downward when its second derivative is negative. For the equation y = (x/3) + 2 + (3/x), the second derivative is negative when x < 0. Therefore, the graph is concave downward for x < 0.
f. Relative maximum: A relative maximum occurs at a point on the graph where the function value is greater than its neighboring points. To find the relative maximum, we need to examine the critical points of the first derivative. The derivative of y = (x/3) + 2 + (3/x) is (1/3) - (3/x^2). Setting this equal to zero gives us (1/3) - (3/x^2) = 0. Solving this equation yields x = ±√9. However, since x = 0 is not defined in the original equation, there are no relative maximum points on the graph.
g. Relative minimum: A relative minimum occurs at a point on the graph where the function value is less than its neighboring points. To find the relative minimum, we again need to examine the critical points of the first derivative. Using the same first derivative obtained in f, we solve (1/3) - (3/x^2) = 0 to find x = ±√9. However, as stated earlier, x = 0 is undefined, so there are no relative minimum points on the graph.
Based on these analyses, the graph of y = (x/3) + 2 + (3/x) is continuous, rising, concave upward for x > 0, and concave downward for x < 0. However, it does not have any relative maximum or relative minimum points.