Q: 4 J in the X F?

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f(x) = 2 * 2 - x + 9 f(-4) = 2 * 2 -(-4) + 9 f(-4) = 4 + 4 + 9 = 17

It is f(x) = 4*3^x.

No, f(x) is not the inverse of f(x).

if f(x) = 4x, then the inverse function g(x) = x/4

When x = 7, f(x) = 3*(4-x) = 3*(-3) = -9

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J. F. X. O'Brien was born in 1828.

J. F. X. O'Brien died in 1905.

void main() { int arr[4][4]; int i,j,a,b,f; printf("\nInput numbers to 4*4 matrix"); for(i=0;i<4;i++) { for(j=0;j<4;j++) { printf("\nKey in the [%d][%d]) value",i+1,j+1); scanf("%d",&arr[i][j]); } } for(i=0;i<4;i++) { for(j=0,f=0;j<4;j++) { if(i!=j&&f==0) continue; a=arr[i][j]; b=arr[j][i]; arr[i][j]=b; arr[j][i]=a; f=1; } } for(i=0;i<4;i++) { for(j=0;j<4;j++) printf("%d ",arr[j][i]); printf("\n"); } }

F. X. J. D'Ambrosio has written: 'Prepare or deter?'

f(x) =6x-4 f(x) = 6(8)-4 f(x) =48-4 f(x) =44

A.) j(a) = a^2 - 2a + 4 B.) j(3) = (3)^2 - 2(3) + 4 = 9 - 6 + 4 = 7 C.) j(x^2) = (x^2)^2 - 2(x^2) + 4 = x^4 - 2x^3 + 4 D.) j(x+3) = (x + 3)^2 - 2(x + 3) + 4 = x^2 +6x + 9 - 2x - 6 + 4 = x^2 + 4x + 7 E.) j(x+h) = (x + h)^2 - 2(x + h) + 4 = x^2 + 2hx + h^2 - 2x - 2h + 4

A.) j(a) = a^2 - 2a + 4 B.) j(3) = (3)^2 - 2(3) + 4 = 9 - 6 + 4 = 7 C.) j(x^2) = (x^2)^2 - 2(x^2) + 4 = x^4 - 2x^3 + 4 D.) j(x+3) = (x + 3)^2 - 2(x + 3) + 4 = x^2 +6x + 9 - 2x - 6 + 4 = x^2 + 4x + 7 E.) j(x+h) = (x + h)^2 - 2(x + h) + 4 = x^2 + 2hx + h^2 - 2x - 2h + 4

A.) j(a) = a^2 - 2a + 4 B.) j(3) = (3)^2 - 2(3) + 4 = 9 - 6 + 4 = 7 C.) j(x^2) = (x^2)^2 - 2(x^2) + 4 = x^4 - 2x^3 + 4 D.) j(x+3) = (x + 3)^2 - 2(x + 3) + 4 = x^2 +6x + 9 - 2x - 6 + 4 = x^2 + 4x + 7 E.) j(x+h) = (x + h)^2 - 2(x + h) + 4 = x^2 + 2hx + h^2 - 2x - 2h + 4

J. F. X. Hoery has written: 'Banjire wis surut' 'Wuludomba pancal panggung'

It is f(x) = -x2 or (-x)2, whichever you intended.

The only way a function can be both even and odd is for it to ignore the input, i.e. for it to be a constant function. e.g. f(x)=4 is both even and odd. An even function is one where f(x)=f(-x), and an odd one is where -f(x)=f(-x). This doesn't make sense. Let's analyze. For a function to be even, f(-x)=f(x). For a function to be odd, f(-x)=-f(x). In this case, f(x)=4, and f(-x)=4. As such, for the first part of the even-odd definition, we have 4=4, which is true, making the function even. However, for the second part of it, we have 4=-4 (f(-x)=4, but -f(x)=-4), which is not true. Therefore constant functions are even because f(-x)=f(x), but not odd because f(-x)!=-f(x).

if f(x) = x² and g(x) = 4 - x, then: fg(x) = (4 - x)² = 16 - 8x + x2 gf(x) = 4 - x² (f+g)(x) = x² + 4 - x (f - g)(x) = x² - 4 + x (fg)(x) = 4x² - x³ (f/g)(x) = x²/4 - x (iff x ≠ 4)