# Two problems dealing with linear approximation and inflection points

• Jan 2nd 2008, 04:21 PM
forkball42
Two problems dealing with linear approximation and inflection points
Okay here are two problems that have my friends and I stumped as we can't figure out a correct answer.

Problem #1:
At the point where x=3 on the curve y = (ax+1)/(x-2) , where a is a constant the slop of the normal is 1/5. Find the value of a.

Its multiple choice and the answers they give us are 1/2, 2, 0, 3, 1/3.
However, I keep getting -5/3 for a.

Problem #2:
The function y=x^4 - 4x^3 + 14 has a horizontal inflection at (A, B). The value of A+B is what?

Yet again it is another multiple choice question. The answers given are -10, 0, -39, 14, or none of these. I have a question though. In class we talk about points of inflection but in this question it talks about horizontal inflection. Is there a difference? We haven't touched base at all in class if it is different.
We found two points of inflection and they were (-2, 2) and (0,14) but we don't know which one is the correct answer if any. Its either 0 or 14 or maybe not if we screwed up in our calculations.

Any help would be appreciated.
• Jan 2nd 2008, 04:37 PM
topsquark
Quote:

Originally Posted by forkball42
Problem #1:
At the point where x=3 on the curve y = (ax+1)/(x-2) , where a is a constant the slop of the normal is 1/5. Find the value of a.

Its multiple choice and the answers they give us are 1/2, 2, 0, 3, 1/3.
However, I keep getting -5/3 for a.

The slope of the tangent to the curve can be found by
$y(x) = \frac{ax + 1}{x - 2}$

$y^{\prime}(x) = \frac{a(x - 2) - (ax + 1)(1)}{(x - 2)^2}$

$y^{\prime}(x) = -\frac{2a + 1}{(x - 2)^2}$

So at x = 3, the tangent to the curve has the value
$y^{\prime}(3) = -\frac{2a + 1}{(3 - 2)^2} = -(2a + 1)$

Now, the normal to the curve at this point is a line with a slope perpendicular to the tangent line. So the slope of the normal line is
$-\frac{1}{-(2a + 1)} = \frac{1}{2a + 1} = \frac{1}{5}$

So solving this I get that a = 2.

-Dan
• Jan 2nd 2008, 04:39 PM
galactus
For #1 the slope is given by $y'=\frac{-(2a+1)}{(x-2)^{2}}$

The normal has slope 1/5, therefore, the slope of the tangent at x=3 must have slope -5.

So, when x=3, $\frac{-(2a+1)}{(3-2)^{2}}=-5$

Solving for a we see a=2
• Jan 2nd 2008, 04:47 PM
topsquark
Quote:

Originally Posted by forkball42
Problem #2:
The function y=x^4 - 4x^3 + 14 has a horizontal inflection at (A, B). The value of A+B is what?

Yet again it is another multiple choice question. The answers given are -10, 0, -39, 14, or none of these. I have a question though. In class we talk about points of inflection but in this question it talks about horizontal inflection. Is there a difference? We haven't touched base at all in class if it is different.
We found two points of inflection and they were (-2, 2) and (0,14) but we don't know which one is the correct answer if any. Its either 0 or 14 or maybe not if we screwed up in our calculations.

This is my best guess after doing a search: A "horizontal inflection point" likely what you would simply call an inflection point. Now, recall that a critical point exists when the first derivative is either 0 or undefined. If the critical point is one where the first derivative is undefined and the second derivative at this point is also undefined, we have what you might call a "vertical inflection point."

In any event, all the inflection points here have a horizontal "tangent" so I would imagine they fall under the category of horizontal inflection points.

$y(x) = x^4 - 4x^3 + 14$

$y^{\prime}(x) = 4x^3 - 12x^2$

So we have critical points at
$y^{\prime}(x) = 4x^3 - 12x^2 = 0 \implies x = 0, 3$

The second derivative is
$y^{\prime \prime}(x) = 12x^2 - 24x$

$y^{\prime \prime}(0) = 0$

$y^{\prime \prime}(3) = 36$

So I get that there is only one inflection point for the function: at x = 0. The y value at x = 0 is y = 14. So the point (A, B) = (0, 14) which implies that A + B = 0 + 14 = 14.

-Dan