Can u pls help me with this question asap
1 answer:
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Answer:
is not possible
Step-by-step explanation:
<u><em>The question in English is</em></u>
we are building a road that links the points a = (12 ,21) and b =(17,23) another point is in c =(3,9) it is possible that a single road allows to join these three points?
we know that
The formula to calculate the slope between two points is equal to
step 1
Find the slope ab
we have
a = (12 ,21) and b =(17,23)
substitute
step 2
Find the slope ac
we have
a = (12 ,21) and c =(3,9)
substitute
simplify
step 3
Compare slopes ab and ac
The slopes are different
That means ----> is not possible that a single road allows to join these three points
The given function is f(x) = x⁶(x-1)⁵
The first derivative is
f'(x) = 6x⁵(x-1)⁵ + 5x⁶(x-1)⁴
= x⁵(x-1)⁴(6x - 6 + 5x)
= x⁵(x-1)⁴(11x - 6)
The critical values are the zeros of f'(x). They are
x = 0, 6/11, and 1.
The critical values indicate that turning points exist for f(x) at the critical points. However, we do not know the nature of the turning points.
Write the first derivative in the form f'(x) = (x-1)⁴(11x⁶ - 6x⁵).
The second derivative is
f''(x) = 4(x-1)³(11x⁶ - 6x⁵) + (x-1)⁴(66x⁵ - 30x⁴)
= (x-1)³(44x⁶ - 24x⁵ + 66x⁶ - 30x⁵ - 66x⁵ + 30x⁴)
= (x-1)³(110x⁶ - 120x⁵ + 30x⁴)
= 10x⁴(x-1)³(11x² - 12x + 3)
The sign of f''(x) at the critical values tell us the nature of the turning point.
f''(0) = 0, therefore a point of inflection exists at x = 0.
f''(6/11) > 0, therefore a local minumum exists at x = 6/11.
f''(1) = 0, therefore a point of inflection exists at x=1.
The graphs shown below confirm these results.
The first derivative is
f'(x) = 6x⁵(x-1)⁵ + 5x⁶(x-1)⁴
= x⁵(x-1)⁴(6x - 6 + 5x)
= x⁵(x-1)⁴(11x - 6)
The critical values are the zeros of f'(x). They are
x = 0, 6/11, and 1.
The critical values indicate that turning points exist for f(x) at the critical points. However, we do not know the nature of the turning points.
Write the first derivative in the form f'(x) = (x-1)⁴(11x⁶ - 6x⁵).
The second derivative is
f''(x) = 4(x-1)³(11x⁶ - 6x⁵) + (x-1)⁴(66x⁵ - 30x⁴)
= (x-1)³(44x⁶ - 24x⁵ + 66x⁶ - 30x⁵ - 66x⁵ + 30x⁴)
= (x-1)³(110x⁶ - 120x⁵ + 30x⁴)
= 10x⁴(x-1)³(11x² - 12x + 3)
The sign of f''(x) at the critical values tell us the nature of the turning point.
f''(0) = 0, therefore a point of inflection exists at x = 0.
f''(6/11) > 0, therefore a local minumum exists at x = 6/11.
f''(1) = 0, therefore a point of inflection exists at x=1.
The graphs shown below confirm these results.