What are the coordinates of the point on the directed line segment from (−4,−10) to (10, -2) that partitions the segment into a
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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.
Answer:
(-8,-10)
Step-by-step explanation:
Rewrite (x+8)2(x+8)² as (x+8)(x+8).
f(x)=3((x+8)(x+8))−10
Expand (x+8) (x+8) using the FOIL Method.
Apply the distributive property.
f(x)=3(x(x+8)+8(x+8))−10
Apply the distributive property.
f(x)=3(x⋅x+x⋅8+8(x+8))−10
Apply the distributive property.
Simplify and combine like terms.
Simplify each term.
Multiply x by x.
f(x)=3(x2+x⋅8+8x+8⋅8)−10
Move 8 to the left of x.
f(x)=3(x2+8⋅x+8x+8⋅8)−10
Multiply 8 by 8.
f(x)=3(x2+8x+8x+64)−10
Add 8x and 8x.
f(x)=3(x2+16x+64)−10
Apply the distributive property.
f(x)=3x2+3(16x)+3⋅64−10
Simplify.
Multiply 16 by 3.
f(x)=3x2+48x+3⋅64−10
Multiply 3 by 64.
f(x)=3x2+48x+192−10
Subtract 10 from 192.
f(x)=3x2+48x+182
The minimum of a quadratic function occurs at x= If a is positive, the minimum value of the function is f (
).
Substitute in the values of aa and b.
x=−
x=-8
Replace the variable x with −8 in the expression.
f(−8)=3(−8)2+48(−8)+182
Y=-10
Therefore, the minimum value is (-8,-10) but if it is asking for just the y-value it would be -10.