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Sagot :
Sure, let's tackle each part of the problem step by step.
### Part (a)
We need to express [tex]\( f(x) = 2x^2 - 16x + 23 \)[/tex] in the form [tex]\( 2(x + a)^2 + b \)[/tex].
To do this, we will complete the square:
1. Factor out the coefficient of [tex]\( x^2 \)[/tex]:
[tex]\[ f(x) = 2(x^2 - 8x) + 23 \][/tex]
2. Complete the square:
[tex]\[ x^2 - 8x \quad \text{becomes} \quad (x - 4)^2 - 16 \][/tex]
Thus:
[tex]\[ 2(x^2 - 8x) = 2((x - 4)^2 - 16) = 2(x - 4)^2 - 32 \][/tex]
3. Add the constant term back:
[tex]\[ f(x) = 2(x - 4)^2 - 32 + 23 = 2(x - 4)^2 - 9 \][/tex]
So, in the form [tex]\( 2(x + a)^2 + b \)[/tex]:
[tex]\[ f(x) = 2(x - 4)^2 - 9 \][/tex]
where [tex]\( a = -4 \)[/tex] and [tex]\( b = -9 \)[/tex].
### Part (b)
To find the range of [tex]\( f(x) = 2(x - 4)^2 - 9 \)[/tex]:
1. Identify the vertex:
The vertex form of a quadratic function [tex]\( a(x - h)^2 + k \)[/tex] shows that the vertex is at [tex]\( (h, k) \)[/tex]. Here, the vertex is at [tex]\( (4, -9) \)[/tex].
2. Determine the direction of the parabola:
Since the coefficient of [tex]\( (x - 4)^2 \)[/tex] is 2 (which is positive), the parabola opens upwards.
Thus, the minimum value of the function is at the vertex.
3. Minimum value:
The minimum value of [tex]\( f(x) \)[/tex] is [tex]\( -9 \)[/tex], and because the parabola opens upwards, the function increases without bound as [tex]\( x \)[/tex] moves away from 4.
4. Range:
[tex]\[ \text{Range of } f: [-9, \infty) \][/tex]
### Part (c)
Let's find an expression for the inverse function [tex]\( f^{-1}(x) \)[/tex]:
Given [tex]\( f(x) = 2(x - 4)^2 - 9 \)[/tex], we need to solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y = 2(x - 4)^2 - 9 \][/tex]
[tex]\[ y + 9 = 2(x - 4)^2 \][/tex]
[tex]\[ \frac{y + 9}{2} = (x - 4)^2 \][/tex]
[tex]\[ x - 4 = \pm \sqrt{\frac{y + 9}{2}} \][/tex]
Since [tex]\( x < 3 \)[/tex], we take the negative branch:
[tex]\[ x - 4 = -\sqrt{\frac{y + 9}{2}} \][/tex]
[tex]\[ x = 4 - \frac{\sqrt{2y + 18}}{2} \][/tex]
Thus, the inverse function is:
[tex]\[ f^{-1}(x) = 4 - \frac{\sqrt{2x + 18}}{2} \][/tex]
### Part (d)
The next function is [tex]\( g(x) = 2x + 4 \)[/tex]. We need to find [tex]\( f(g(x)) \)[/tex] and simplify it.
Let's substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ g(x) = 2x + 4 \][/tex]
[tex]\[ f(g(x)) = f(2x + 4) = 2(2x + 4)^2 - 16(2x + 4) + 23 \][/tex]
First, simplify [tex]\( (2x + 4)^2 \)[/tex]:
[tex]\[ (2x + 4)^2 = 4x^2 + 16x + 16 \][/tex]
Now substitute:
[tex]\[ f(2x + 4) = 2(4x^2 + 16x + 16) - 16(2x + 4) + 23 \][/tex]
[tex]\[ = 8x^2 + 32x + 32 - 32x - 64 + 23 \][/tex]
[tex]\[ = 8x^2 + 32 - 64 + 23 \][/tex]
[tex]\[ = 8x^2 - 9 \][/tex]
Thus, the simplified expression for [tex]\( f(g(x)) \)[/tex] is:
[tex]\[ f(g(x)) = 8x^2 - 9 \][/tex]
So, in summary:
(a) [tex]\( f(x) = 2(x - 4)^2 - 9 \)[/tex]
(b) The range of [tex]\( f \)[/tex] is [tex]\( [-9, \infty) \)[/tex]
(c) [tex]\( f^{-1}(x) = 4 - \frac{\sqrt{2x + 18}}{2} \)[/tex]
(d) [tex]\( f(g(x)) = 8x^2 - 9 \)[/tex]
### Part (a)
We need to express [tex]\( f(x) = 2x^2 - 16x + 23 \)[/tex] in the form [tex]\( 2(x + a)^2 + b \)[/tex].
To do this, we will complete the square:
1. Factor out the coefficient of [tex]\( x^2 \)[/tex]:
[tex]\[ f(x) = 2(x^2 - 8x) + 23 \][/tex]
2. Complete the square:
[tex]\[ x^2 - 8x \quad \text{becomes} \quad (x - 4)^2 - 16 \][/tex]
Thus:
[tex]\[ 2(x^2 - 8x) = 2((x - 4)^2 - 16) = 2(x - 4)^2 - 32 \][/tex]
3. Add the constant term back:
[tex]\[ f(x) = 2(x - 4)^2 - 32 + 23 = 2(x - 4)^2 - 9 \][/tex]
So, in the form [tex]\( 2(x + a)^2 + b \)[/tex]:
[tex]\[ f(x) = 2(x - 4)^2 - 9 \][/tex]
where [tex]\( a = -4 \)[/tex] and [tex]\( b = -9 \)[/tex].
### Part (b)
To find the range of [tex]\( f(x) = 2(x - 4)^2 - 9 \)[/tex]:
1. Identify the vertex:
The vertex form of a quadratic function [tex]\( a(x - h)^2 + k \)[/tex] shows that the vertex is at [tex]\( (h, k) \)[/tex]. Here, the vertex is at [tex]\( (4, -9) \)[/tex].
2. Determine the direction of the parabola:
Since the coefficient of [tex]\( (x - 4)^2 \)[/tex] is 2 (which is positive), the parabola opens upwards.
Thus, the minimum value of the function is at the vertex.
3. Minimum value:
The minimum value of [tex]\( f(x) \)[/tex] is [tex]\( -9 \)[/tex], and because the parabola opens upwards, the function increases without bound as [tex]\( x \)[/tex] moves away from 4.
4. Range:
[tex]\[ \text{Range of } f: [-9, \infty) \][/tex]
### Part (c)
Let's find an expression for the inverse function [tex]\( f^{-1}(x) \)[/tex]:
Given [tex]\( f(x) = 2(x - 4)^2 - 9 \)[/tex], we need to solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y = 2(x - 4)^2 - 9 \][/tex]
[tex]\[ y + 9 = 2(x - 4)^2 \][/tex]
[tex]\[ \frac{y + 9}{2} = (x - 4)^2 \][/tex]
[tex]\[ x - 4 = \pm \sqrt{\frac{y + 9}{2}} \][/tex]
Since [tex]\( x < 3 \)[/tex], we take the negative branch:
[tex]\[ x - 4 = -\sqrt{\frac{y + 9}{2}} \][/tex]
[tex]\[ x = 4 - \frac{\sqrt{2y + 18}}{2} \][/tex]
Thus, the inverse function is:
[tex]\[ f^{-1}(x) = 4 - \frac{\sqrt{2x + 18}}{2} \][/tex]
### Part (d)
The next function is [tex]\( g(x) = 2x + 4 \)[/tex]. We need to find [tex]\( f(g(x)) \)[/tex] and simplify it.
Let's substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ g(x) = 2x + 4 \][/tex]
[tex]\[ f(g(x)) = f(2x + 4) = 2(2x + 4)^2 - 16(2x + 4) + 23 \][/tex]
First, simplify [tex]\( (2x + 4)^2 \)[/tex]:
[tex]\[ (2x + 4)^2 = 4x^2 + 16x + 16 \][/tex]
Now substitute:
[tex]\[ f(2x + 4) = 2(4x^2 + 16x + 16) - 16(2x + 4) + 23 \][/tex]
[tex]\[ = 8x^2 + 32x + 32 - 32x - 64 + 23 \][/tex]
[tex]\[ = 8x^2 + 32 - 64 + 23 \][/tex]
[tex]\[ = 8x^2 - 9 \][/tex]
Thus, the simplified expression for [tex]\( f(g(x)) \)[/tex] is:
[tex]\[ f(g(x)) = 8x^2 - 9 \][/tex]
So, in summary:
(a) [tex]\( f(x) = 2(x - 4)^2 - 9 \)[/tex]
(b) The range of [tex]\( f \)[/tex] is [tex]\( [-9, \infty) \)[/tex]
(c) [tex]\( f^{-1}(x) = 4 - \frac{\sqrt{2x + 18}}{2} \)[/tex]
(d) [tex]\( f(g(x)) = 8x^2 - 9 \)[/tex]
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