Answer:
Step-by-step explanation:
We want an exponential function that goes through the two points (0, 8) and (2, 200).
Since a point is (0, 8), this means that y = 8 when x = 0. Therefore:
Simplify:
So we now have:
Likewise, the point (2, 200) tells us that y = 200 when x = 2. Therefore:
Solve for b. Dividing both sides by 8 yields:
Thus:
Hence, our exponential function is:
<em>Response-</em>
<em>ABC is absolutely a right triangle</em>
<em>we can use the pythagorean theorem to solve this</em>
<h3><em>Definitions</em></h3>Right Triangle - <em>A right triangle or right-angled triangle, or more formally an orthogonal triangle, is a triangle in which one angle is a right angle. The relation between the sides and other angles of the right triangle is the basis for trigonometry. The side opposite to the right angle is called the hypotenuse.</em>
Pythagorean Theorem - <em>In mathematics, the Pythagorean theorem, or Pythagoras' theorem, is a fundamental relation in Euclidean geometry among the three sides of a right triangle. It states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides.</em>
<em>_________________</em>
<em>To use to the Pythagorean Theorem it is</em>
For our equation would be
would be
<em>_________________</em>
<h2><em>Solve</em></h2>Since we got this is a right triangle since it's what we had before.
Answer:
(a) Decreasing on (0, 1) and increasing on (1, ∞)
(b) Local minimum at (1, 0)
(c) No inflection point; concave up on (0, ∞)
Step-by-step explanation:
ƒ(x) = x² - x – lnx
(a) Intervals in which ƒ(x) is increasing and decreasing.
Step 1. Find the zeros of the first derivative of the function
ƒ'(x) = 2x – 1 - 1/x = 0
2x² - x -1 = 0
( x - 1) (2x + 1) = 0
x = 1 or x = -½
We reject the negative root, because the argument of lnx cannot be negative.
There is one zero at (1, 0). This is your critical point.
Step 2. Apply the first derivative test.
Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.
(1) x = ½
ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1
ƒ'(x) < 0 so the function is decreasing on (0, 1).
(2) x = 2
ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½ = ⁵/₂
ƒ'(x) > 0 so the function is increasing on (1, ∞).
(b) Local extremum
ƒ(x) is decreasing when x < 1 and increasing when x >1.
Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.
(c) Inflection point
(1) Set the second derivative equal to zero
ƒ''(x) = 2 + 2/x² = 0
x² + 2 = 0
x² = -2
There is no inflection point.
(2). Concavity
Apply the second derivative test on either side of the extremum.
The function is concave up on (0, ∞).
