Answer :
Answer:
To find the largest value of the function \( f(x) = 2(a - x)(x + \sqrt{x^2 + b}) \), we need to determine the maximum value of the function over its entire domain.
Let's first expand the function:
\( f(x) = 2(a - x)(x + \sqrt{x^2 + b}) \)
\( f(x) = 2ax + 2\sqrt{x^2 + b} - 2x^2 - 2x\sqrt{x^2 + b} \)
\( f(x) = 2ax + 2\sqrt{x^2 + b} - 2x^2 - 2x\sqrt{x^2 + b} \)
Now, to find the maximum value of the function, we need to take the derivative of the function with respect to x and set it equal to zero to find the critical points:
\( f'(x) = 2a - 4x - \frac{2x + b}{\sqrt{x^2 + b}} \)
Setting \( f'(x) = 0 \) and solving for x:
\( 2a - 4x - \frac{2x + b}{\sqrt{x^2 + b}} = 0 \)
\( 2a - 4x = \frac{2x + b}{\sqrt{x^2 + b}} \)
\( (2a - 4x)^2 = (2x + b)^2 \)
\( 4a^2 - 16ax + 16x^2 = 4x^2 + 4bx + b^2 \)
\( 12x^2 - 16ax - 4bx + b^2 - 4a^2 = 0 \)
Now, we can solve this quadratic equation to find the critical points. Once we have the critical points, we can evaluate the function at those points and at the endpoints of the domain to determine the largest value of the function.
Step-by-step explanation:
Step-by-step explanation:
f(x) - 2(a - x)(x + sqrt(x ^ 2 + b ^ 2)), x \in R f' * (x) = 2(a - x)(1 + (2x)/(2sqrt(x ^ 2 + b ^ 2))) - 2(x + sqrt(x ^ 2 + b ^ 2)) f' * (x) = 2(x + sqrt(x ^ 2 + b ^ 2))((a - x)/(sqrt(x ^ 2 + b ^ 2)) - 1)
For maxima/minima:
f' * (x) = 0
2(x + sqrt(x ^ 2 + b ^ 2))((a - x)/(sqrt(x ^ 2 + b ^ 2)) - 1) = 0 a - x = sqrt(x ^ 2 + b ^ 2)
(a - x) ^ 2 = x ^ 2 + b ^ 2
a ^ x + x ^ 2 - 2ax = x ^ 2 + b ^ 2
x = (a ^ 2 - b ^ 2)/(2a)
Then we check:
F a-b 2a <0- maxima at x = a-b 2a tax of x)=2(a- a^ 2 -b^ 2 2a )( (a ^ 2 - b ^ 2)/(2a) + sqrt(((a ^ 2 - b ^ 2)/(2a)) ^ 2 + b ^ 2) , a²+b² a²-b² a²+b² 2a = 2((a ^ 2 + b ^ 3)/(2a))(a) -a²+b