The value of the derivative at the maximum or minimum for a continuous function must be zero.
<h3>What happens with the derivative at the maximum of minimum?</h3>
So, remember that the derivative at a given value gives the slope of a tangent line to the curve at that point.
Now, also remember that maximums or minimums are points where the behavior of the curve changes (it stops going up and starts going down or things like that).
If you draw the tangent line to these points, you will see that you end with horizontal lines. And the slope of a horizontal line is zero.
So we conclude that the value of the derivative at the maximum or minimum for a continuous function must be zero.
If you want to learn more about maximums and minimums, you can read:
brainly.com/question/24701109
X^2+x-30 -- find two numbers that add to one and multiply to -30, which would be 6 and -5. so now you have x^2+6x-5x-30. factor out x and 5 to get x(x+6)-5(x+6). so you get (x+6)*(x-5)
(a) Using the table, give the values fo rthe inverse
1) original table of values:
x 1 2 3 4 5
f(x) 0 1 1 5 3
2) The inverse of the function is obtained by exchanging x and f(x), this is:
( x, f(x) ) → ( f(x), x)
3) So, the table of values of the inverse of the given function is:
x 0 1 1 5 3
f⁻¹ (x) 0 1 2 3 4
(b) Is the inverse a function?
No, the inverse is not a function, since the table of the inverse shows that the x -value 1 has two different images.
This ambigüity is opposite to the definition of a function, which requires that any input value has only one output. For that reason, the inverse is not a function. You cannot tell whether the image of 1 is 1 or 2, because both are images of the same value.
Answer:
so one side right?
Step-by-step explanation:
Let x = number of months.
Let y = total number of fish in pond.
6% of 3,000 = 180 fish
We can use an equation to help us with this problem.
y = 3,000 + 180x - 400x. Combine like terms.
y = 3,000 - 220x. Plug 9 in for x.
y = 3,000 - 220(9). Solve for y
y = 1,020.
There are 1,020 fish in the pond after 9 months.
