Question about maximum values (no working out needed)?

Question

I recently got this question in a coursework:

Find the maximum value of f(x) = 4x^2 - x^4 + 4.
For what value(s) of x does this occur.
That was the full question, it didn't state any desired method to use.

So I got: dy/dx = 8x - 4x^3

worked out that when dy/dx = 0, x=0 or +/- sqrt(2)

then I got: d^2y/dx^2 = 8 - 12x^2

so when x = 0, d^2y/dx^2 = +8, ie a minimum.

when x = +/-sqrt(2), d^2y/dx^2 = -16, ie maximum.

so I put x = +/- sqrt(2) back into the original equation to get 8.

My final answer was the maximum is 8, and this occurs when x = +/- sqrt(2).

But I only got 2 out of 4 marks. He/she (I'm not sure who marked it) put a note next to my second derivative saying that I have to evaluate them. I don't know what this means, I have evaluated the answer and surely I don't have to evaluate f(0) if I know it's a minimum.

What are your thoughts?

Answer 1

Sure looks like a 4 out of 4 response to me! I used to
teach this stuff.
I suggest you request a face-to-face meeting with your
teacher, asking for an explanation of the comment, i.e.
"what did I miss"? The answer, I hope, should be a
correction of grade.
If you do this, I'd love to know the outcome.

Answer 2

it looks ok to me
I assumt the function is defined on the real line

Maybe he/she wanted to see how you got -16 or
maybe he wanted to write:
d^2x/dx^2(sqrt(2)) = -16 because d^2x/dx^2 is a function, not a number

But anyhow I would have given you 4 marks. That error is minor .

edit: it was also better if you wrote x is maximum(minimum) because the second derivative evaluated in x is positive(negative)

Answer 3

Let x^2 = y. Then -(y^2 -4y -4) = -( (y - 2)^2 - 8).

Minimum value of (y-2)^2 is zero, when y =2. So function has maximum value of 8, when x is +/- square root of 2.

Simpler, at least.

Answer 4

you have showed that f'(x) =0 and f "(x) <0

A local maximum was found.

It's very unreasonable not to give you full mark

Answer 5

go get your marks back