Given graph of the function $f(x)=x^3+ax^2+bx+c$ What is the $x$ value at the local minimum point?

Question

This is a problem from a timed exam, so I prefer approaches that lead to answer quickly.

Graph of the function $f(x)=x^3+ax^2+bx+c$ is as follow. What is the $x$ value at the local minimum point?

$1)\frac12\qquad\qquad2)2\qquad\qquad3)\frac32\qquad\qquad4)3$

Here is my approach,

Suppose we have minima at $x_0$. We have,$$f(0)=4\Rightarrow c=4$$

$$f'(x)=3x^2+2ax+b\qquad\text{Since $f'(0)=0\rightarrow b=0$}$$Hence $f'(x)=x(3x+2a)$, So $x_0=-\frac{2a}3$. We have $f(x)=x^3+ax^2+4$. $$f(-\frac{2a}3)=0\Rightarrow -\frac{8a^3}{27}+\frac{4a^3}9+4=0\Rightarrow a=-3$$ Finally $x_0=\frac{-2a}3=2$.

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Although one can get the answer with this approach in one or two minutes, I'm looking for quicker ways to solve the problem. Actually, first time I tried to solve the problem, I noticed that we can write $f(x)=(x+m)(x-n)^2$ where $m,n\in \mathbb{R}^+$. According to graph of function and noting that $f(0)=mn^2=4$, I can guess it the function is $f(x)=(x+1)(x-2)^2$. But assuming I'm in the exam, at this point should I quickly mark the answer and go to the next question or this approach is unreliable and I got the correct answer with a bit of luck?!

Answer 1

From the shape of the graph you have two roots, call them $p$ for the single on the left and $q$ for the double root on the right, so the equation can be factorised as $(x-p)(x-q)^2=0$. This results in $$ x^3-(p+2q)x^2+(2pq+q^2)x-pq^2=0 $$ The intercept gives us $-pq^2=4$ and the local maximum at $x=0$ gives us $2pq+q^2=0$ which gives us $q=-2p$, thus $p=-1$ and $q=2$.

Answer 2

Not a great deal faster, but you could also use the fact that the local extrema in the cubic function are symmetric about its inflection point, given by $ \ f''(x) \ = \ 6x + 2a \ = \ 0 $ $ \Rightarrow \ x_{infl} \ = \ -\frac{a}{3} \ \ . $ Since the local maximum is on the $ \ y-$axis (if the diagram is to be at all trusted), then the local minimum is at $ \ x_{min} \ = \ -\frac{2a}{3} \ \ . $ The function value at the inflection point is $ \ f\left(-\frac{a}{3} \right) \ = \ -\frac{a^3}{27} + \frac{a^3}{9} - \frac{ab}{3} + 4 \ = \ 2 \ \ . $ You will want to use your observation that $ \ f'(0) \ = \ 0 \ \Rightarrow \ b \ = \ 0 \ \ , $ so we have $ \ \frac{2·a^3}{27} \ = \ -2 \ \Rightarrow \ a^3 \ = \ -27 $ $ \Rightarrow \ x_{min} \ = \ -\frac{2·[-3]}{3} \ = \ +2 \ \ . $

Your educated guess that $ \ x^3 - 3x^2 + 4 \ = \ (x + 1)·(x - 2)^2 \ $ is correct; acting on that during the exam depends on how much of a gambler you are: Viete tells you that $ \ c \ = \ 4 \ = \ -r·s^2 \ \ , $ so $ \ r \ = \ -1 \ \ , \ \ s \ = \ +2 \ $ would be worth the risk if you're desperate for time...

ADDENDUM: It might also be noted that once you see that $ \ c \ = \ 4 \ \ , $ the Rational Zeroes Theorem lets you "reject" choices (3) and (4) (now... does $ \ \frac{1}{2} \ $ seems plausible...?).

Answer 3

Suppose that a cubic function has critical points at $x = r$ and $x = s$. IOW, its derivative is zero at these points. Then

$$f'(x) = k(x - r)(x - s)$$ $$f'(x) = k(x^2 - (r+s)x + rs)$$ $$f(x) = k(\frac{1}{3}x^3 - \frac{1}{2}(r + s)x^2 + rsx) + c$$

But you're given that the $x^3$ coefficient is 1, so $k = 3$.

$$f(x) = x^3 - \frac{3}{2}(r + s)x^2 + 3rsx + c$$

You're also given $r = 0$ and $c = 4$ at the other local extremum, so:

$$f(x) = x^3 - \frac{3}{2}sx^2 + 4$$

But at this local minimum point, $x = s$ and $f(x) = 0$, so:

$$x^3 - \frac{3}{2}x^3 + 4 = 0$$ $$-\frac{1}{2}x^3 = -4$$ $$x^3 = 8$$ $$x = 2$$

Answer 4

Let $s$ and $t$ denote the $x$-intercepts.

Then

$$ f(x)=(x-s)(x-t)^2 $$

Since we know $f(0)=4$ that gives $-st^2=4$ so $s=-\frac{4}{t^2}$

Next, we have

$$f^\prime(x)=(x-t)^2+2(x-s)(x-t)$$

Since $f^\prime(0)=0$ we have $t^2+2st=0$. Since we already have $s=-\frac{4}{t^2}$ we substitute to get $t^3-8=0$ giving $t=2$.

Notice that all you really need from the equation given for $f(x)$ is the fact that the leading coefficient of the polynomial is $1$. Everything else is given by the graph of $f$.

Answer 5

It may be a little faster to use the symmetry about the inflection point. Say the inflection point is at $(h,2)$, so $$f(x)=\left.x(x^2-v^2)\right\rvert_{\text{shifted}}=(x-h)\left((x-h)^2-v^2\right)+2$$

$f(0)=4$, so $$-h\left(h^2-v^2\right)=2\tag{!}$$

And $$f'(x)=3(x-h)^2-v^2$$ and then $f'(0)=0$, so $$3h^2-v^2=0$$

Now you know $v^2=3h^2$, so return to (!):

$$-h\left(-2h^2\right)=2$$

This implies $h=1$. So the positive root of $f$ is at $x=2h=2$.