# Thread: [SOLVED] Interpreting Curve Sketching Conditions

1. ## [SOLVED] Interpreting Curve Sketching Conditions

Interpret each piece of information, so that you can be able to sketch the graph of a function $\displaystyle f$ that satisfies the following conditions:

.(a) $\displaystyle f(0) = 0$

.(b) $\displaystyle f''(x) > 0$, . $\displaystyle x \neq 0$

.(c) $\displaystyle \mathop {\lim }\limits_{x \to 0^-} f'(x) = \infty$

.(c) $\displaystyle \mathop {\lim }\limits_{x \to 0^+} f'(x) = -\infty$

.(d) $\displaystyle \mathop {\lim }\limits_{x \to -\infty} f(x) = -\infty$

.(d) $\displaystyle \mathop {\lim }\limits_{x \to \infty} f(x) = \infty$

.

My interpretations are as follows:

.(a) y-intersect at (0,0)

.(b) Concave up for all values of $\displaystyle x$, except at $\displaystyle x = 0$ (possibly an asymptote there?)

.(c) As you approach 0 from the left, the slope approaches infinity and so the function gets higher.

.(c) As you approach 0 from the right, the slope approaches negative infinity and so the function gets lower.

.(d) As you approach negative infinity (left), the function gets lower and lower,

.(d) As you approach positive infinity (right), the function gets higher and higher.

I'm not to sure about them though, and it seems like (b) and (c) are contradictory (if C is true, that would mean concave down to the right of 0).

Did I mis-interpret these conditions or is there a flaw in the question?

UPDATE - Correct Answer is as follows: (thanks to Soroban)(changes in red)

.(a) y-intersect at (0,0)

.(b) Concave up for all values of x, except at x = 0 [ which is where g''(x) = 0 ].

.(c) As you approach 0 from the left, the slope is positive and it is getting more and more vertical (approaching infinity), and so the function is getting higher.

.(c) As you approach 0 from the right, the slope is negative and it is getting more and more vertical (approaching infinity), and so the function is getting higher.

.(d) As you approach negative infinity (left), the function gets lower and lower,

.(d) As you approach positive infinity (right), the function gets higher and higher.

Therefore, the function is indeed concave up from the right side of 0.

2. Hello, Some_One!

Most of your interpretations are correct.
Some clarification is needed on two of them. *

Interpret each piece of information, so that you can be able to sketch the graph
of a function $\displaystyle f$ that satisfies the following conditions:

$\displaystyle (a)\;\;f(0) = 0$

$\displaystyle (b)\;\;f''(x) > 0,\;\;x \neq 0$

$\displaystyle (c)\;\;\lim_{x \to 0^-} f'(x) \:=\: \infty$

$\displaystyle (c)\;\;\lim_{x \to 0^+} f'(x) \:=\: -\infty$

$\displaystyle (d)\;\;\lim_{x \to -\infty} f(x) \:=\: -\infty$

$\displaystyle (d)\;\;\lim_{x \to \infty} f(x) \:=\: \infty$

My interpretations are as follows:

(a) y-intercept at (0,0)

(b) Concave up for all values of $\displaystyle x$, except at $\displaystyle x = 0$ (possibly an asymptote there?)

(c) As you approach 0 from the left, the slope approaches infinity
and so the function gets higher. *

(c) As you approach 0 from the right, the slope approaches negative infinity
and so the function gets lower. *

If we approach from the right, the function gets higher.

(d) As you approach negative infinity (left), the function gets lower and lower.

(d) As you approach positive infinity (right), the function gets higher and higher.

"As we approach 0 from the left, the slope approaches infinity" . . . true!
. . And the curve is rising . . . but not infinitely.
If the slope is becoming infinite, the curve is getting "more vertical".
. . It may not soar off into the sky.

Consider the lower half of a circle. Start at 6 o'clock and move CCW to 3 o'clock. See it?

I suspect that the graph looks like this:
Code:
|
|               *
|
|              *
- - - - - - - - * - - - - - - * - - -
|*          *
*|  *     *
* |     *
*   |
*      |
*          |
*              |
|

3. Originally Posted by Soroban
Hello, Some_One!

Most of your interpretations are correct.
Some clarification is needed on two of them. *

"As we approach 0 from the left, the slope approaches infinity" . . . true!
. . And the curve is rising . . . but not infinitely.
If the slope is becoming infinite, the curve is getting "more vertical".
. . It may not soar off into the sky.

Consider the lower half of a circle. Start at 6 o'clock and move CCW to 3 o'clock. See it?

I suspect that the graph looks like this:
Code:
|
|               *
|
|              *
- - - - - - - - * - - - - - - * - - -
|*          *
*|  *     *
* |     *
*   |
*      |
*          |
*              |
|
Thank you very, very much!

I think I get it now! So "negative $\displaystyle \infty$" means that the slope is negative, and it is approaching infinity (getting more vertical). Therefore, it is indeed concave up.

It would have been going downwards if it would be "positive $\displaystyle \infty$" (as x approaches 0 from the right).