Curve Sketching

Tags

Cegep1

Mathematics

Word count

471 words

Reading time

3 minutes

To sketch the curve of a function $f$:

1. Find the domain of $f$.
2. Find the x-intercepts and the y-intercept if possible.
3. Find the horizontal and vertical asymptotes of $f$ if possible.
4. Find the intervals of monotonicity of $f$.
5. (Optional) find the relative extrema of $f$.
6. Find the intervals of concavity of $f$.
7. Find the inflection points of $f$.
8. Sketch the graph.
   1. Draw a Cartesian plane.
   2. Plot the x-intercepts, the y-intercept, the H.A., the V.A., the relative extrema, and the inflection points.
   3. Plot $f(x)$ from left to right.
      • Consider the monotonicity and concavity of $f$.
      • Consider the behaviour of $f$ from left and right of V.A.
      • Consider the side of $f$ relative to the H.A.

Examples

Given

$$\begin{array}{rl}f(x)& =(5-2x)x^{\frac{2}{3}}\\ f^{\prime }(x)& =\frac{10(1-x)}{3x^{\frac{1}{3}}}\\ f^{″}(x)& =\frac{-10(1+2x)}{9x^{\frac{4}{3}}}\end{array}$$

sketch the graph of $f$.

Domain

$f$ is defined $\mathrm{\forall }x\in \mathbb{R}$.

Intercepts

x-intercepts:

$$\begin{array}{rl}f(x)& =0\\ ⟹(5-2x)x^{\frac{2}{3}}& =0\\ 5-2x=0& \text{ or }{x}^{\frac{2}{3}}=0\\ x=\frac{5}{2}& x=0\end{array}$$

y-intercept:

$$f(0)=(5-2\cdot 0)0^{\frac{2}{3}}=0$$

Asymptotes

V.A.:

Since $domf=x\in \mathbb{R}$, there are no V.A.

H.A.:

$$\begin{array}{rl}\underset{x\to \pm \mathrm{\infty }}{lim}f(x)& =\underset{x\to \pm \mathrm{\infty }}{lim}(5-2x)x^{\frac{2}{3}}\\ & =\underset{x\to \pm \mathrm{\infty }}{lim}(5-2x)\underset{x\to \pm \mathrm{\infty }}{lim}{x}^{\frac{2}{3}}\\ & =\mp \mathrm{\infty }\cdot \mathrm{\infty }\\ & =\mp \mathrm{\infty }\end{array}$$

Therefore, there are no H.A.

Intervals of monotonicity and concavity

$$\begin{array}{rl}{f}^{\prime }(x)& =0\\ 10(1-x)& =0\\ 1-x& =0\\ x& =1\end{array}$$

$f^{\prime }(x)$ is undefined if

$$\begin{array}{rl}3x^{\frac{1}{3}}& \mathrm{\infty }=0\\ x^{\frac{1}{3}}& =0\\ x& =0\end{array}$$

Similarly,

$$\begin{array}{rl}{f}^{″}(x)& =0\\ -10(1+2x)& =0\\ 1+2x& =0\\ x& =-\frac{1}{2}\end{array}$$

$f^{″}(x)$ is undefined if

$$\begin{array}{rl}9x^{\frac{4}{3}}& =0\\ x^{\frac{4}{3}}& =0\\ x& =0\end{array}$$

So,

$-\mathrm{\infty }$	$-\frac{1}{2}$	0	1	$\mathrm{\infty }$
$f^{\prime }(x)$	-	-	+	-
$f^{″}(x)$	+	-	-	-
$f(x)$	dec dec	inc dec	dec inc	inc dec

Relative extrema

$x=0$ is a relative minimum.
$x=1$ is a relative maximum:

$$\begin{array}{rl}f(1)& =(5-2)\cdot 1^{\frac{2}{3}}\\ & =3\end{array}$$

Inflection points

$x=0$ is not an inflection point since the concavity doesn't change.
$x=-\frac{1}{2}$ is an inflection point:

$$\begin{array}{rl}f(-\frac{1}{2})& =(5-2(-\frac{1}{2})){(-\frac{1}{2})}^{\frac{2}{3}}\\ & =\frac{6}{\sqrt[3]{4}}\end{array}$$

Curve sketch

Curve Sketching 1

Contributors

Ajitani Hifumi

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