minimum value of f(x)

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as08998078

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(9) Let vector  a→ = (2, 3) \displaystyle \, \overrightarrow{a}\, =\, (2,\, 3)\, and  b→ = (−1, 4). \displaystyle \, \overrightarrow{b}\, =\, (-1,\, 4).\, When  x→ − a→ = b→ − x→, \displaystyle \, \overrightarrow{x}\, -\, \overrightarrow{a}\, =\, \overrightarrow{b}\, -\, \overrightarrow{x},\, then  x→ = (12, 72)\displaystyle \, \overrightarrow{x}\, =\, \boxed{\left(\dfrac{1}{2},\, \dfrac{7}{2}\right)}

(10) Let  f(x) = −x3 + 6x2 − 9x + 1.\displaystyle \, f(x)\, =\, -x^3\, +\, 6x^2\, -\, 9x\, +\, 1.

(i) The derivative  f′(x) = −3x2 + 12 − 1\displaystyle \, f'(x)\, =\, \boxed{-3x^2\, +\, 12\, -\, 1}

(ii) Under  0 ≤ x ≤ 3, \displaystyle \, 0\, \leq\, x\, \leq\, 3,\, the minimum value of  f(x) \displaystyle \, f(x)\, is: _________

2: There is a parabola  A : y = x2 − 4x − 5.\displaystyle \, A\, :\, y\, =\, x^2\, -\, 4x\, -\, 5.

(1) The coordinate of the vertex of the parabola A is.....


In (ii) I don't know how to find the minimum value of f(x). Please help me.
(In the box, it is answer)
 
Last edited by a moderator:
as08998078 said:
(10) Let  f(x) = −x3 + 6x2 − 9x + 1.\displaystyle \, f(x)\, =\, -x^3\, +\, 6x^2\, -\, 9x\, +\, 1.

(i) The derivative  f′(x) = −3x2 + 12 − 1\displaystyle \, f'(x)\, =\, \boxed{-3x^2\, +\, 12\, -\, 1}

(ii) Under  0 ≤ x ≤ 3, \displaystyle \, 0\, \leq\, x\, \leq\, 3,\, the minimum value of  f(x) \displaystyle \, f(x)\, is: _________


In (ii) I don't know how to find the minimum value of f(x).
Click to expand...
What have you learned about derivatives and their relation to max/min points? What do you know about finding the x-intercepts (that is, the zeroes) of quadratics? ;)
 
Since the domain is restricted, there could be global max/min at the end points.
 
Last edited by a moderator:
as08998078 said:
(9) Let vector  a→ = (2, 3) \displaystyle \, \overrightarrow{a}\, =\, (2,\, 3)\, and  b→ = (−1, 4). \displaystyle \, \overrightarrow{b}\, =\, (-1,\, 4).\, When  x→ − a→ = b→ − x→, \displaystyle \, \overrightarrow{x}\, -\, \overrightarrow{a}\, =\, \overrightarrow{b}\, -\, \overrightarrow{x},\, then  x→ = (12, 72)\displaystyle \, \overrightarrow{x}\, =\, \boxed{\left(\dfrac{1}{2},\, \dfrac{7}{2}\right)}

(10) Let  f(x) = −x3 + 6x2 − 9x + 1.\displaystyle \, f(x)\, =\, -x^3\, +\, 6x^2\, -\, 9x\, +\, 1.

(i) The derivative  f′(x) = −3x2 + 12 − 1\displaystyle \, f'(x)\, =\, \boxed{-3x^2\, +\, 12\, -\, 1} ..... Incorrect  f′(x) = −3x2 + 12x − 9\displaystyle \, f'(x)\, =\, {-3x^2\, +\, 12x\, -\, 9}

(ii) Under  0 ≤ x ≤ 3, \displaystyle \, 0\, \leq\, x\, \leq\, 3,\, the minimum value of  f(x) \displaystyle \, f(x)\, is: _________

2: There is a parabola  A : y = x2 − 4x − 5.\displaystyle \, A\, :\, y\, =\, x^2\, -\, 4x\, -\, 5.

(1) The coordinate of the vertex of the parabola A is.....


In (ii) I don't know how to find the minimum value of f(x). Please help me.
(In the box, it is answer)
Click to expand...
.
 
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