For the real valued functions f(x) = x-4/x-5 and g(x) = 4x+17, find the composition fog and specify its domain using interval no
tation.
(F o g)(x)=
Domain of f o g:
(F o g)(x)=
Domain of f o g:
1 answer:
Answer:
(f o g)(x) =
Domain : (-∞, -3) ∪ (-3, ∞)
Step-by-step explanation:
Given : f(x) =
g(x) = 4x + 17
We have to find the value of (f o g)(x).
Since (f o g)(x) = f[g(x)]
=
=
For the Domain of the function (f o g)(x),
Given function is defined for (4x + 12) ≠ 0
4x ≠ -12
x ≠ -3
Therefore, Domain of the function will be : (-∞, -3) ∪ (-3, ∞)
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Answer:
The dimension of the box is 17.66 in by 7.66 in by 2.67 in.
Therefore the volume of the box is 361.19 .
Step-by-step explanation:
Given that the dimensions of a cardboard is 23 in by 13 in.
Let the side of the square be x in.
Then the length of the box= (23-2x) in
and the width of the box =(13-2x) in
and height = x in.
The volume of the box is = length ×width × height
=[(23-2x)(13-2x)x]
=(299x-72x² +4)
∴V=299x-72x² +4x³
Differentiating with respect x
V'= 299-144x+12x²
Again differentiating with respect x
V''= -144+24x
To find the dimensions, we set V'=0
∴299-144x+12x²=0
Applying Sridharacharya formula that is the solution of a quadratic equation ax²+bx+c is
Here a=12, b=-144 , c=299
If we take x=9.33 in, then the width of the box [13-(2×9.33)] will negative.
∴x = 2.67 in
If at x = 2.67, V''<0 , then the volume of the box will be maximum or V''>0 then volume of the box will be minimum.
Therefore at x = 2.67, the volume of the box maximum.
The length of the box =[23-(2×2.67)] in
=17.66 in
The width of the box =[13-(2×2.67)] in
=7.66 in
The height of the box= 2.67 in
The dimension of the box is 17.66 in by 7.66 in by 2.67 in.
Therefore the volume of the box is =(17.66×7.66×2.67)
=361.19