What is coordinates of the y intercept of x+5=0
1 answer:
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We can apply some rules backwards
first
f'(x)=
(x-4)(-2x+4)
-2x^2+12x-16
we know that
so therefor maybe
-2x^2=rnx^{n-1}
2=n-1
n=3
rn=-2
3r=-2
r=-2/3
one is
second part
12x
12x=
x^1, 1=n-1
n=2
rn=12
2r=12
r=6
is the second bit
last part
-16
-16x^0=
0=n-1
n=1
rn=-16
1r=-16
r=-16
-16x^1
so therfor f(x)=
first
f'(x)=
(x-4)(-2x+4)
-2x^2+12x-16
we know that
so therefor maybe
-2x^2=rnx^{n-1}
2=n-1
n=3
rn=-2
3r=-2
r=-2/3
one is
second part
12x
12x=
x^1, 1=n-1
n=2
rn=12
2r=12
r=6
last part
-16
-16x^0=
0=n-1
n=1
rn=-16
1r=-16
r=-16
-16x^1
so therfor f(x)=
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)
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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