The given polynomial function has 1 relative minimum and 1 relative maximum.
<h3>What are the relative minimum and relative maximum?</h3>
- The relative minimum is the point on the graph where the y-coordinate has the minimum value.
- The relative maximum is the point on the graph where the y-coordinate has the maximum value.
- To determine the maximum and the minimum values of a function, the given function is derivated(since the maximum or minimum is obtained at slope = 0)
<h3>Calculation:</h3>
The given function is
f(x) = 2x³ - 2x² + 1
derivating the above function,
f'(x) = 6x² - 4x
At slope = 0, f'(x) = 0 (for maximum and minimum values)
⇒ 6x² - 4x = 0
⇒ 2x(3x - 2) = 0
2x = 0 or 3x - 2 = 0
∴ x = 0 or x = 2/3
Then the y-coordinates are calculated by substituting these x values in the given function,
when x = 0;
f(0) = 2(0)³ - 2(0)² + 1 = 1
So, the point is (0, 1)
when x = 2/3;
f(2/3) = 2(2/3)³ - 2(2/3)² + 1 = 19/27
So, the point is (2/3, 19/27)
Since y = 1 is the largest value, the point (0, 1) is the relative maximum for the given function.
So, y = 19/27 is the smallest value, the point (2/3, 19/27) is the relative minimum for the given function.
Thus, option A is correct.
Learn more about the relative minimum and maximum here:
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<h3>
Answer: f(x) = x + 13 </h3>
This is the same as y = x+13
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Explanation:
Let's find the slope
I'll use the first two rows as the (x1,y1) and (x2,y2) points
m = (y2-y1)/(x2-x1)
m = (19-18)/(6-5)
m = 1/1
m = 1
The slope is 1.
Now apply the point slope formula and solve for y
y - y1 = m(x - x1)
y - 18 = 1(x - 5)
y - 18 = x - 5
y = x-5 + 18
y = x + 13
f(x) = x + 13 is the final answer
As a check, note how something like x = 5 leads to...
f(x) = x+13
f(5) = 5+13 ... replace x with 5
f(5) = 18
We see that x = 5 leads to f(x) = 18. That verifies the first row. I'll let you check the remaining three rows.
The equation y = x+13 has slope 1 and y intercept 13.
Answer:
4:10, 2:5 16:40
Step-by-step explanation:
just multiply or divide
Answer:
a1= 1 q= −sinx , dla |q| <1 , ta suma jest zbieżna
a1 1
S=
=
1 −q 1+sinx
w mianowniku podobnie: a1=1 , q= sinx , dla | sinx| <1
1
S=
1 −sinx
i mamy równanie:
1
1+sinx
= tg2x
1
1− sinx
Step-by-step explanation: