Answer:
8.36
Step-by-step explanation:
Write in Slope Intercept Form: (0,4) (3,18)
2 answers:
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Answer:
You could use a piece wise function to simulate the speed someone is running at a given time in minutes.
for s = 9 mph
for s = 0 mph
for s = 6 mph
Step-by-step explanation:
I can't do graphs right now but there's an idea
The given function is
f(x) = x - ln(8x), on the interval [1/2, 2].
The derivative of f is
f'(x) = 1 - 1/x
The second derivative is
f''(x) = 1/x²
A local maximum or minimum occurs when f'(x) = 0.
That is,
1 - 1/x = 0 => 1/x = 1 => x =1.
When x = 1, f'' = 1 (positive).
Therefore f(x) is minimum when x=1.
The minimum value is
f(1) = 1 - ln(8) = -1.079
The maximum value of f occurs either at x = 1/2 or at x = 2.
f(1/2) = 1/2 - ln(4) = -0.886
f(2) = 2 - ln(16) = -0.773
The maximum value of f is
f(2) = 2 - ln(16) = -0.773
A graph of f(x) confirms the results.
Answer:
Minimum value = 1 - ln(8)
Maximum value = 2 - ln(16)
f(x) = x - ln(8x), on the interval [1/2, 2].
The derivative of f is
f'(x) = 1 - 1/x
The second derivative is
f''(x) = 1/x²
A local maximum or minimum occurs when f'(x) = 0.
That is,
1 - 1/x = 0 => 1/x = 1 => x =1.
When x = 1, f'' = 1 (positive).
Therefore f(x) is minimum when x=1.
The minimum value is
f(1) = 1 - ln(8) = -1.079
The maximum value of f occurs either at x = 1/2 or at x = 2.
f(1/2) = 1/2 - ln(4) = -0.886
f(2) = 2 - ln(16) = -0.773
The maximum value of f is
f(2) = 2 - ln(16) = -0.773
A graph of f(x) confirms the results.
Answer:
Minimum value = 1 - ln(8)
Maximum value = 2 - ln(16)