Solve for the variable using Trig ratio.
1 answer:
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Answer:
Where 0 < x < 3
The location of the local minimum, is (2, 0)
The location of the local maximum is at (0, 16)
Step-by-step explanation:
The given function is f(x) = (x + 2)⁴
The range of the minimum = 0 < x < 3
At a local minimum/maximum values, we have;
∴ (-x + 2)³ = 0
x = 2
When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2
We have, f(2) = (-2 + 2)⁴ = 0
The location of the local minimum, is (2, 0)
Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3
When x = 0, -x + 2 = 0 + 2 = 2
Similarly, we have;
-x + 2 = 1, when x = 1
-x + 2 = 0, when x = 2
-x + 2 = -1, when x = 3
Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16
The location of the local maximum is at (0, 16).
Answer:
<em>The two figures are similar as the ratio of corresponding distances and slope of corresponding lines are EQUAL.</em>
Step-by-step explanation:
Noting the coordinates of points,
C(-1,6), D(3,4) , E(2,3) , F(3,0), G(1,-2) , A(0,1) , B(-2,2)
J(4.5,3.5), K(6.5,2.5), L(6,2), M(6.5,0.5) , N(5.5,-0.5) , H(5,1) , I(4,1.5)
using the distance formula =
we find that,
CD= and JK=
thus, ratio of CD to JK is 2.
Similarly find ratio of DE to KL and so on, the ratio remains same.
Further one can even comment on the slopes of corresponding lines been equal.
Thus the figures turn out to be similiar to each other