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A-Level Turning Points

The curve C has equation y=4x^2 + (5x-1)/x

a. Find f'(x)

For this I've got 8x + x^-2

b. Find the x-coordinate of the stationary point of C

Now this is where it all goes downhill because I got x=0 but looking on Desmos says that x=0 is a vertical asymtote.

c. Determine whether this stationary point is a maximum or a minimum

To do this I know you have to feed the x-coordinate back through the f''(x) function, and if it's negative it's a maximum and if it's positive it's a minimum.
Original post by Retsek
The curve C has equation y=4x^2 + (5x-1)/x

a. Find f'(x)

For this I've got 8x + x^-2

b. Find the x-coordinate of the stationary point of C

Now this is where it all goes downhill because I got x=0 but looking on Desmos says that x=0 is a vertical asymtote.

c. Determine whether this stationary point is a maximum or a minimum

To do this I know you have to feed the x-coordinate back through the f''(x) function, and if it's negative it's a maximum and if it's positive it's a minimum.


How are you getting x=0?
Reply 2
Original post by NotNotBatman
How are you getting x=0?


Because I got f'(x) = 8x + x^-2
and the stationary point is when f(x) = 0
and that only happens when x = 0 no?

EDIT: Okay no never mind 1/0 is undefined my bad
(edited 6 years ago)
Original post by Retsek
Because I got f'(x) = 8x + x^-2
and the stationary point is when f(x) = 0
and that only happens when x = 0 no?

EDIT: Okay no never mind 1/0 is undefined my bad


4x2+5x1x4x2+5xx1x...4x^2+\frac{5x-1}{x}\equiv 4x^2+\frac{5x}{x}-\frac{1}{x}\equiv...
Reply 4
Original post by BuryMathsTutor
4x2+5x1x4x2+5xx1x...4x^2+\frac{5x-1}{x}\equiv 4x^2+\frac{5x}{x}-\frac{1}{x}\equiv...


Original post by NotNotBatman
How are you getting x=0?


24098709_146582205967817_1740815358_n.jpg
Original post by Retsek
Because I got f'(x) = 8x + x^-2
and the stationary point is when f(x) = 0
and that only happens when x = 0 no?

EDIT: Okay no never mind 1/0 is undefined my bad


Stationary point is when f(x)=0f'(x)=0, so you want to show that f(x)=0f'(x)=0 has no solutions.
Original post by Retsek
24098709_146582205967817_1740815358_n.jpg


Sorry, I misread your work. Please ignore my post.

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