Could you help me with interpreting this solution?

Question

Problem: The polynomial $f(x)=ax^2+bx+c$ has a local maximum on the positive $y$ axis. Decide the conditions that must be true for the coefficients $a$, $b$ and $c$. Show how you do this.

Solution: $$f'(x)=2ax+b$$ $$f'(0)=2a\cdot 0+b=0\, , \;b=0 \rightarrow f'(x)=2ax.$$ The maximum demands that $f'(x)>0$ for $x$ negative, which means that $a$ is also negative, $a<0$. The positive $y$ axis gives us that $f(0)>0\rightarrow c>0$.

I do not understand this way of reasoning, could you please help me understand it - for example add more details to it, explain why it is as it is/ stands as it stands?

Answer 1

• If the maximum lies on $y$ axis then it's $x$ coordinate must be $0$ which is $x=0$
• If a value is a extreme point then it's derivative is zero,plugging in $x=0$ we get $f'(0)=0$
• Now in order for $f(0)$ to be a maximum the function must be increasing to a maximum then decreasing,here we need for $x<0$ to have that $f'(x)>0$ or differently put that $f(x)$ is increasing and for $x>0$ to have that $f'(x)<0$ is decreasing which is exactly when $a<0$
• The last step is that $f(0)=c>0$ because it must be on the positive part of $y$ axis

Answer 2

The language seems just slightly (but not very obtuse). Let's see what we can do:

"The second-grade polynomial"

I'm not sure what "second-grade" means. I assume it means it has degree 2 or is a quadratic.

"$ f(x)=ax^2+bx+c$ has a local maximum"

I hope I don't have to explain that local maximum means that at some value of $x=x_0$ that means that $f(x_0)$ is larger than those values of $f(x)$ for $x$ slightly less than $x_0$ and for $x$ slightly more than $x_0$.

More technically, I guess I could say, there is an $r>0$ so that for all $x$ so that $x_0 - r < x < x_0 + r$ that $f(x_0) > f(x)$. But I hope, I don't need to explain that.

Also I hope you are familiar that for $x_0$ to be the site of a local extrema that $f'(x_0)=0$. (Otherwise $f(x)$ is increasing or decreasing at $x=x_0$ and it can't be an extreme value.)

" on the positive y-axis."

That means $x_0 = 0$ and $f(x) > 0$.

So we need $f(0) > 0$. And we need $f'(0) = 0$.

"Solution:

$"f′(x)=2ax+b$".

That's straightforward, isn't it? $f(x) = ax^2 + bx + c$ so $f'(x) = 2ax + b$.

$f′(0)=2a⋅0+b=0,b=0$"

Because $f'(0) = 2(0)x + b = b$ and we know to have a local maximum at $x=0$ that $f'(0) = 0$ we know $b=0$.

"$f′(x)=2ax$".

$f'(x) = 2ax + b$ and $b = 0$ so $f'(x) = 2ax$.

"The maximum demands that f′(x)>0 for x negative,"

Okay, that's a weird way of putting it. But if $f(0)$ is a maximum then and the values $x < 0$ $f(x)$ must be increasing. So $f'(x) > 0$ for $x< 0$. Likewise at the $x > 0$ $f(x)$ must be decreasing. So $f'(x) < 0$ for $x >0$.

" which means that a is also negative, a<0"

If $x < 0$ and $f'(x) = 2a(x) > 0$ then this must mean $a < 0$.

I'd prefer to state it as $f$ achieves i) a maximum at $x = x_0$ if $f'(x_0) = 0$ and $f''(x) < 0$, ii) a minimum at $x = x_0$ if $f'(x_0) = 0$ and $f''(x_0) > 0$. iii) a saddle point at $x=x_9$ if $f'(x_0) = 0$ and$f''(x_0) = 0$.

Then I'd note $f''(x) = 2a< 0$ means $a < 0$.

"The positive y-axis gives us that f(0)>0→c>0"

This is $f(0) > 0$. So $f(0) = a0^2 + b*0 + c > 0$. So $c > 0$. So

$f(x) = zx^2 + bx + c$ has a positive local maximum at $x=0$ if $a < 0; b= 0; c>0$