Answer with explanation:
→→→Function 1
f(x)= - x²+ 8 x -15
Differentiating once , to obtain Maximum or minimum of the function
f'(x)= - 2 x + 8
Put,f'(x)=0
-2 x+ 8=0
2 x=8
Dividing both sides by , 2, we get
x=4
Double differentiating the function
f"(x)= -2, which is negative.
Showing that function attains maximum at ,x=4.
Now,f(4)=-4²+ 8× 4-15
= -16 +32 -15
= -31 +32
=1
→→→Function 2:
f(x) = −x² + 2 x − 3
Differentiating once , to obtain Maximum or minimum of the function
f'(x)= -2 x +2
Put,f'(x)=0
-2 x +2=0
2 x=2
Dividing both sides by , 2, we get
x=1
Double differentiating the function,gives
f"(x)= -2 ,which is negative.
Showing that function attains maximum at ,x=1.
f(1)= -1²+2 ×1 -3
= -1 +2 -3
= -4 +2
= -2
⇒⇒⇒Function 1 has the larger maximum.
Let, the length = l
Width = l/2
Area = l * w = l * l/2 = l²/2
So, Your Final Answer would be A = l²/2
Hope this helps!
I dont know math well but try to look it up
5,13,15 and more!
hope that helps!!!!
so, no breaks on the denominator thus, the LCD will be all three then.
![\bf \cfrac{3}{x-2}+\cfrac{4}{x-1}-\cfrac{1}{x}\implies \cfrac{[3x(x-1)]+[4x(x-2)]-[(x-2)(x-1)]}{x(x-2)(x-1)} \\\\\\ \cfrac{[3x^2-3x]+[4x^2-8x]-[x^2-3x+2]}{x(x-2)(x-1)} \\\\\\ \cfrac{3x^2-3x+4x^2-8x-x^2+3x-2}{x(x-2)(x-1)} \\\\\\ \cfrac{6x^2-8x-2}{x(x-2)(x-1)}\implies \cfrac{2(x^2-4x-1)}{x(x-2)(x-1)}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B3%7D%7Bx-2%7D%2B%5Ccfrac%7B4%7D%7Bx-1%7D-%5Ccfrac%7B1%7D%7Bx%7D%5Cimplies%20%5Ccfrac%7B%5B3x%28x-1%29%5D%2B%5B4x%28x-2%29%5D-%5B%28x-2%29%28x-1%29%5D%7D%7Bx%28x-2%29%28x-1%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%5B3x%5E2-3x%5D%2B%5B4x%5E2-8x%5D-%5Bx%5E2-3x%2B2%5D%7D%7Bx%28x-2%29%28x-1%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B3x%5E2-3x%2B4x%5E2-8x-x%5E2%2B3x-2%7D%7Bx%28x-2%29%28x-1%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B6x%5E2-8x-2%7D%7Bx%28x-2%29%28x-1%29%7D%5Cimplies%20%5Ccfrac%7B2%28x%5E2-4x-1%29%7D%7Bx%28x-2%29%28x-1%29%7D)
