Find b such that f(x)=-4x^2+bx+3 has a maximum value of 50.
The derivative f'(x) of -4x²+bx+3 is the gradient of f(x), which is -8x+b.
At x=50 the function takes a maximum value, so the gradient must be zero, so -8x+b=0 at x=50, which means that b=400.
Check the above: f(x) = -4x² + 400x + 3, so f'(x) = -8x + 400 = 0, so x = 50.
To find the value of b such that the function f(x) = -4x^2 + bx + 3 has a maximum value of 50, we need to determine the vertex of the parabolic function.
The vertex of a parabola in the form f(x) = ax^2 + bx + c can be found by using the formula x = -b/2a, where a and b are coefficients of x^2 and x, respectively.
In our case, a = -4 and the y-coordinate of the vertex is 50. Plugging these values into the formula, we have:
x = -b / (2 * -4)
50 = -4 * x^2 + b * x + 3
To isolate b, let's rearrange the equation:
4x^2 - bx + 53 = 0
Since the vertex is the maximum point, the quadratic equation has only one solution for x, which means it is a perfect square trinomial.
To represent our quadratic equation as a perfect square trinomial, we complete the square:
4(x^2 - (b/4x) + (b^2/16)) + 53 - (b^2/4) = 0
Simplifying further, we have:
x^2 - (b/4x) + (b^2/16) + 53/4 - (b^2/4) = 0
To make it a perfect square trinomial, the constant term (-b^2/4) should be equal to the square of half the coefficient of x.
Therefore, (b^2/4) = (1/2 * -b/4)^2
b^2/4 = (b^2/16)
16b^2 = 4b^2
12b^2 = 0
b = 0
Therefore, the value of b that makes the function have a maximum value of 50 is 0.
To find the value of b such that the quadratic function f(x) = -4x^2 + bx + 3 has a maximum value of 50, we need to determine the vertex of the parabola represented by the function.
The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the coordinates (-b/2a, f(-b/2a)). In this case, a = -4.
Since we want to find the maximum value of the function, the vertex should lie at the peak of the parabola. Let's denote the value of b that satisfies this condition as b_max.
Now, the x-coordinate of the vertex is given by -b_max/2a. In this case, a = -4. Substituting the given value of 50 for f(-b_max/2a), we get:
50 = -4(-b_max/2(-4))^2 + b_max(-b_max/2(-4)) + 3
Simplifying this equation, we have:
50 = -1/2b_max^2 - b_max^2/2 + 3
Combining like terms, we get:
50 = -3/2b_max^2/2 + 3
To solve for b_max, we subtract 3 from both sides to isolate the term:
50 - 3 = -3/2b_max^2/2
47 = -3/2b_max^2/2
Next, we multiply both sides by 2 to eliminate the fraction:
94 = -3b_max^2
Dividing both sides by -3, we get:
-94/3 = b_max^2
Finally, taking the square root of both sides of the equation, we find:
± √(-94/3) = b_max
However, since we are considering a real-world scenario here, the maximum value of a quadratic function can only be achieved when it opens downward, which means the coefficient of x^2 should be negative. Therefore, we discard the positive square root solution, leaving us with the negative square root:
b_max = - √(-94/3)
Hence, the value of b that ensures f(x) = -4x^2 + bx + 3 has a maximum value of 50 is b_max = - √(-94/3).