How the graph of $y=x^3-5x^2+7x-3$ may be sketched without any electronic aid?

Question

Also, the max point is $(1,0)$, but the graph later extends upto infinity. So how can $(1,0)$ be the maximum point. Same for the minimum point.

Answer 1

1. Differentiate the function and set it to zero to find all maxima's and minima's.

2. Take 30 well placed points on x axis which will cover all the critical points.

3. Work out value of y for these points (may take 40 mins by hand or 5 mins by calculator)

4. Plot approximately and extrapolate.

5. To minimize error, check algebraically the x and y intercepts and match them while plotting.

6. This method will give you a very accurate graph.

Done

Answer 2

Hint:  by inspection (or by the rational root theorem) $x=1$ is a root of both the function and its derivative, which yields the factorizations:

$$ y = (x-1)^2(x-3) \\ y' = (x-1)(3x-7) $$

Once the zeros and the (local) extrema of the function are determined, and the limits at $\pm\infty$ are obvious, the graph can be sketched with some confidence.

Answer 3

For $x\rightarrow+\infty$ we have $y\rightarrow+\infty$, which says that $y$ have no a maximal value.

For $x\rightarrow-\infty$ we have $y\rightarrow-\infty$, which says that $y$ have no a minimal value.

But $y$ has a local maximum and a local minimum.

Indeed, $y=(x-1)^2(x-3)$, which says that $y\leq0$ for all $0\leq x\leq3$,

which says that $(1,0)$ is a maximum point by definition of the maximum point.

Now, let $1<x<3$.

Hence, by AM-GM $$y=(x-1)^2(x-3)=-\frac{1}{2}(x-1)(x-1)(6-2x)\geq$$ $$\geq-\frac{1}{2}\left(\frac{x-1+x-1+6-2x}{3}\right)^3=-\frac{32}{27}.$$ The equality occurs for $x-1=6-2x$, which is $x=\frac{7}{3}$.

Thus, $\left(\frac{7}{3},-\frac{32}{27}\right)$ is a minimum point by definition of the minimum point.

Answer 4

Well, you can work as follows:

1. Let $f:\mathbb{R}\to\mathbb{R}$ be a function with $f(x)=x^3-5x^2+7x-3$.
2. Find the roots of the function, or the intervals they are located. In our case: $$f(x)=0\Leftrightarrow=x^3-5x^2+7x-3=0\Leftrightarrow...\Leftrightarrow x=1\mbox{ or }x=3$$ So, $f$ meets the $x-$axis on $A=(1,0)$ and $B=(3,0)$.
3. Find the point that $f$ meets $y-$axis. This is always, since $f$ is a function, the point $C=(0,f(0))$, when $f(0)$ is defined. In our case: $$f(0)=-3$$ so $C=(0,-3)$.
4. Check $f$'s continuity and differentiability. In our case, $f$ is a polynomial and, hence, it is infinitely continuously differentiable, so we have to be careful and plot a really smooth graph - no edges etc.
5. Find $f$'s monotonicity and extrema. For this purpose, we first calculate $f'$: $$f'(x)=3x^2-10x+7=(x-1)(3x-7)$$ So $$f'(x)=0\Leftrightarrow\dots\Leftrightarrow x=1\mbox{ or }x=\frac{7}{3}$$ which means that critical point are $x_1=1$ and $x_2=\frac{7}{3}$. So, we can easily get the following table: $$\begin{array}{c|ccccccc} x & -\infty & & 1 & & \frac{7}{3} & & +\infty\\ \hline f'(x) & & + & 0 & - & 0 & + & \\ \hline f(x) & & \nearrow & \text{max} & \searrow & \text{min} & \nearrow & \\ \end{array}$$
6. Find $f$'s convexity and inflection point(s). For this purpose, we calculate $f''$: $$f''(x)=6x-10$$ So $$f''(x)=0\Leftrightarrow x=\frac{5}{3}$$ which means that $x_0=\frac{5}{3}$ is the only possible inflection point. So, we can easily get the following table: $$\begin{array}{c|ccccc} x & -\infty & & \frac{5}{3} & & +\infty\\ \hline f''(x) & & - & 0 & + & \\ \hline f(x) & & \curvearrowright & \text{i.p.} & \circlearrowleft & \\ \end{array}$$ (I could not find an appropriate arrow for convexity and rotatebox did not seem to work).
7. Check $f$'s asymptotical behaviour. Well, since $f$ is a polynomial of degree 3 we need not check anything.
8. Check $f$'s limits on the boundaries of its domain. In our case, we have: $$\lim_{x\to+\infty}f(x)=\dots=+\infty$$ $$\lim_{x\to-\infty}f(x)=\dots=-\infty$$
9. Plot $f$ with all that in mind.

Hope this helped! :)