Sketch the graphs of the functions - Calculus question

[Mary4ever]

Homework Statement

Sketch the graphs of the functions. Indicate intervals on which the function is increasing, decreasing, concave up, or concave down; indicate relative maximum points, relative minimum points, points of inflection, horizontal asymptotes, vertical asymptotes, symmetry, and those intercepts that can be obtained conveniently:

Homework Equations

y=x/((2x-1)^3)

The Attempt at a Solution

I have a solution, but it involves a graph, which I cannot put here

 

[SammyS]

Homework Statement

Sketch the graphs of the functions. Indicate intervals on which the function is increasing, decreasing, concave up, or concave down; indicate relative maximum points, relative minimum points, points of inflection, horizontal asymptotes, vertical asymptotes, symmetry, and those intercepts that can be obtained conveniently:

Homework Equations

y=x/((2x-1)^3)

The Attempt at a Solution

I have a solution, but it involves a graph, which I cannot put here

Hello Mary4ever. Welcome to PF !

That makes it difficult to say whether your solution is correct or not.

Can you give your results for: intervals on which the function is increasing, decreasing, concave up, or concave down; indicate relative maximum points, relative minimum points, points of inflection, horizontal asymptotes, vertical asymptotes, symmetry, and those intercepts that can be obtained conveniently ?

 

[Mary4ever]

For relative maxima and minima,

f'(x) = 0 gives e^x = e^{-x}, which happens when x=0, only.

f''(0) =f(0) = 1

and so x=0 is a minimum. It is also an absolute minimum, because f is increasing for
x>0 and decreasing for x<0.

f''(x) > 0 for all x and so concave up everywhere, no points of inflection.

Please let me know if this is correct, and please let me know the asymptotes and how to sketch the graph. Thank you

 

[Mark44]

For relative maxima and minima,

f'(x) = 0 gives e^x = e^{-x}, which happens when x=0, only.

Where does e^x come into this problem?

Your function is f(x) = x/(2x - 1)³.

f''(0) =f(0) = 1

and so x=0 is a minimum. It is also an absolute minimum, because f is increasing for
x>0 and decreasing for x<0.

f''(x) > 0 for all x and so concave up everywhere, no points of inflection.

Please let me know if this is correct, and please let me know the asymptotes and how to sketch the graph. Thank you