The height of a hill, h(x), in a painting can be written as a
2 answers:
Answer:
First option:
Step-by-step explanation:
<h3> The complete exercise is: "The height of a hill, h(x), in a painting can be written as a function of x, the distance from the left side of the painting. Both h(x) and x are measured in inches
You have the following function provided in the exercise:
You know that represents the height of the hill (in inches) and "x" represents the distance from the left side of the painting (in inches)
Knowing that you can determine that, if the painting 3 inches from the left side of the picture, the value of "x" is the following:
Therefore, you need to find the value of when
in order to solve this exercise.
So, the next step is to substitute into the function:
And finally, you must evaluate in order to find .
You get that this is:
You might be interested in
A) 
is continuous, so 
is also continuous. This means if we were to integrate 
, the same constant of integration would apply across its entire domain. Over 
, we have 
. This means that
For to be continuous, we need the limit as 
to match 
. This means we must have
Now, over
, we have 
, so 
, which means 
.
b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,
c) is increasing when 
, and concave upward when 
, i.e. when is also increasing.
We have
over the intervals 
and 
. We can additionally see that 
only on and .
d) Inflection points occur when
, and at such a point, to either side the sign of the second derivative 
changes. We see this happening at 
, for which 
, and to the left of we have decreasing, then increasing along the other side.
We also have along the interval 
, but even if we were to allow an entire interval as a "site of inflection", we can see that to either side, so concavity would not change.
For
Now, over
b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,
c)
We have
d) Inflection points occur when
We also have
Answer:
x = 1.9
Step-by-step explanation:
use SOH-CAH-TOA
