The absolute minimum = -2√2.
The absolute maximum= 4.5
Consider f(t)=t√9-t on the interval (1,3].
Find the critical points: Find f'(t)=0.
f"(t) = 0
√9-t² d/dt t + t d/dt √9-t²=0
√9-t² + t/2√9-t² (-2t)=0
9-t²-t²/√9-t²=0
9-2t²=0
9=2t², t²=9/2, t=±3/√2
since -3/√2∉ (1,3].
Therefore, the critical point in the interval (1,3] is t= 3/√2.
Find the value of the function at t=1, 3/√2,3 to find the absolute maximum and minimum.
f(-1)=-1√9-1²
= -√8 , =-2√2
f(3/√2)= 3/√2 √9-(3/√2)²
= 3/√2 √9-9/2
=3/√2 √9/2
=9/2 = 4.5
f(3)= 3√9-3²
= 3(0)
=0
The absolute maximum is 4.5 and the absolute minimum is -2√2.
The absolute maximum point is the point at which the function reaches the maximum possible value. Similarly, the absolute minimum point is the point at which the function takes the smallest possible value.
A relative maximum or minimum occurs at an inflection point on the curve. The absolute minimum and maximum values are the corresponding values over the full range of the function. That is, the absolute minimum and maximum values are bounded by the function's domain.
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Answer:
B) 25x^2 – 100
Step-by-step explanation:
Step-by-step explanation:
y — yı = m(x – xı) equation of a line
Points A(1,3) and B(2,1)
Slope (m) = (y2 – y1) / (x2 – x1)
(1 –3) / (2 – 1)
–2/1
–2
Thus, y –3 = –2 (x – 1)
y – 3 = –2x + 2
y –3 + 3 = –2x + 2 + 3
y = –2x + 5
Answer:
The first equation should be multiplied by 9 and the second equation by −4
Step-by-step explanation:
Given the simultaneous equation
First Equation: 5x − 4y = 28
Second equation: 3x - 9y = 30
In order to eliminate y, we must make the coefficient of x in both expression to be equal.
To do that the first equation should be multiplied by 9 (negative value of the coefficient of y in equation 2)and the second equation by -4( (coefficient of y in equation 1)
The two equations are
y=7x+5
y=4x+25
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