Let f(x) = −x2 + ax + b.
Determine the constants a and b such that f has a relative maximum at x = 2 and the relative maximum value is 13.
Well, it seems like f(x) is getting really picky with its relative maximum! I hope it knows what it's looking for because sometimes even I have trouble finding my maximum clown potential.
Since we want a relative maximum at x = 2, we need to find the derivative of f(x) and set it equal to zero. So, let's find that derivative!
f'(x) = -2x + a
To find the relative maximum, we need to set the derivative equal to zero:
-2x + a = 0
Solving for x, we get:
x = a/2
Now, we know that the relative maximum value is 13, so f(2) = 13. Substituting x = 2 into the original equation, we get:
f(2) = -2^2 + a(2) + b = 13
Simplifying that equation, we have:
-4 + 2a + b = 13
Now, we have two equations:
a/2 = 2 (from setting the derivative equal to zero)
-4 + 2a + b = 13 (from substituting x = 2 into the original equation)
Solving these equations, we find:
a = 4
b = 15
So, the constants a and b that satisfy the conditions are a = 4 and b = 15. Now let's hope f(x) appreciates our efforts to find its relative maximum!
To determine the values of constants a and b for the function f(x) = -x^2 + ax + b, such that f has a relative maximum at x = 2 and the relative maximum value is 13, we need to follow the steps below:
Step 1: Differentiate the function f(x) with respect to x to find its derivative.
The derivative of f(x) is given by f'(x) = -2x + a.
Step 2: Set the derivative f'(x) equal to 0 to find critical points.
Since f has a relative maximum at x = 2, we know that at this point, f'(x) = 0.
Thus, we set -2x + a = 0 and solve for x:
-2(2) + a = 0
-4 + a = 0
a = 4.
Step 3: Substitute the value of a into the original function f(x).
Now that we have found the value of a, we can substitute it into the original function f(x) and solve for b by using the given information that the relative maximum value is 13.
f(x) = -x^2 + ax + b
f(x) = -x^2 + 4x + b
To find the value of b, we substitute x = 2 and f(x) = 13 into the equation:
13 = -2^2 + 4(2) + b
13 = -4 + 8 + b
13 = 4 + b
b = 9.
Therefore, the values of the constants a and b that satisfy the given conditions are a = 4 and b = 9, respectively. So, the function f(x) = -x^2 + 4x + 9 has a relative maximum at x = 2 with a value of 13.
To determine the constants a and b that satisfy the given conditions, we can use the properties of relative maximum.
The first condition states that f has a relative maximum at x = 2. A relative maximum occurs when the derivative of the function changes from positive to negative at that point. Thus, we need to find the derivative of the function f(x) and evaluate it at x = 2.
1. Find the derivative of f(x):
f'(x) = d/dx(-x^2 + ax + b) = -2x + a
2. Evaluate the derivative at x = 2:
f'(2) = -2(2) + a = -4 + a
The second condition states that the relative maximum value is 13. A relative maximum occurs at the point (x, y) where y is the maximum value. In this case, the maximum value is 13, so we can set f(2) = 13.
3. Substitute x = 2 in f(x):
f(2) = -2^2 + a(2) + b
f(2) = -4 + 2a + b
Since the relative maximum occurs at x = 2, we have the following two equations:
1. f'(2) = -4 + a = 0 (because the derivative changes sign at the relative maximum)
2. f(2) = -4 + 2a + b = 13 (because the maximum value is 13)
Solve these two equations simultaneously to find the values of a and b.
From equation 1, we have: a = 4.
Substitute the value of a in equation 2:
-4 + 2(4) + b = 13
-4 + 8 + b = 13
4 + b = 13
b = 9
Therefore, the constants a and b that satisfy the given conditions are a = 4 and b = 9.
Correction: b=9
f(x)=-x^2 +4x+b
if f(2)=13=-2^2+4(2)+b
then f(2)=13=-4+8+b
17=8+b
9=b
f'(x) = -2x+a
if f'=0 at x=2, then a=4
f(x) = -x^2 + 4x + b
f(2) = 13 = -4+8+b
b=5
f(x) = -x^2 + 4x + 5 = -(x+1)(x-5)
vertex is at x=(5-1)/2 = 2 as desired