Given the function f(x) = ax^3 + bx^2 + 3x - 2,
and f(2) = 10 and f ' (-1) = 14,
solve for A and B.
Ah, math problems! They're like a puzzle at a clown convention. Alright, let's solve this one together.
First, let's substitute the given values into our function:
f(2) = 10:
10 = a(2)^3 + b(2)^2 + 3(2) - 2
Simplifying, we get:
10 = 8a + 4b + 6
And now, let's look at the second given condition:
f'(-1) = 14:
(The prime notation represents the derivative)
14 = 3a(-1)^2 + 2b(-1) + 3
Simplifying once again:
14 = 3a - 2b + 3
Now, we have two equations with two unknowns. Let's solve them simultaneously.
From the first equation, we rearrange it to isolate a:
8a + 4b = 4
From the second equation, we rearrange it to isolate a:
-2b + 3a = 11
Now, let's solve for a:
From the first equation, we get:
8a = 4 - 4b
a = (4 - 4b) / 8
a = (1 - b) / 2
Now we substitute this expression for a into the second equation:
-2b + 3((1 - b) / 2) = 11
-2b + (3 - 3b) / 2 = 11
-4b + 3 - 3b = 22
-7b = 19
b = -19/7
Now that we've found the value for b, we can substitute it back into the first equation to solve for a:
8a + 4(-19/7) = 4
8a - (76/7) = 4
8a = (4 + 76/7)
8a = (28 + 76)/7
8a = 104/7
a = 13/7
So, the solution is a = 13/7 and b = -19/7.
I hope these calculations didn't leave you feeling too divided!
To solve for A and B, we can use the given information and the properties of derivatives.
Step 1: Find the derivative of f(x).
Taking the derivative of f(x) with respect to x, we get:
f'(x) = 3ax^2 + 2bx + 3
Step 2: Plug in the values f(2) = 10 and f'(-1) = 14.
Using the given value for f(2), we have:
f(2) = a(2^3) + b(2^2) + 3(2) - 2 = 8a + 4b + 6 - 2 = 8a + 4b + 4 = 10
Using the given value for f'(-1), we have:
f'(-1) = 3a(-1)^2 + 2b(-1) + 3 = 3a - 2b + 3 = 14
Step 3: Solve the system of equations.
We now have the following system of equations:
8a + 4b + 4 = 10
3a - 2b + 3 = 14
Simplifying the equations, we get:
8a + 4b = 6 (Equation 1)
3a - 2b = 11 (Equation 2)
We can solve this system of equations by eliminating one variable.
To eliminate b, we multiply Equation 2 by 2 and add it to Equation 1:
(2)(3a - 2b) + (8a + 4b) = 22 + 6
6a - 4b + 8a + 4b = 28
14a = 28
a = 28/14
a = 2
Step 4: Substitute the value of a back into either equation to solve for b.
Using Equation 1:
8(2) + 4b = 6
16 + 4b = 6
4b = 6 - 16
4b = -10
b = -10/4
b = -5/2 or -2.5
Therefore, the values of A and B that satisfy the given conditions are A = 2 and B = -2.5.
To solve for A and B in the function f(x) = ax^3 + bx^2 + 3x - 2, given the conditions f(2) = 10 and f'(-1) = 14, we can use the concepts of evaluating a function and finding the derivative.
1. Start with the function f(x) = ax^3 + bx^2 + 3x - 2.
2. To find f(2), substitute x = 2 into the function and set it equal to 10.
f(2) = a(2)^3 + b(2)^2 + 3(2) - 2 = 8a + 4b + 6 - 2 = 8a + 4b + 4 = 10.
3. Simplify the equation obtained in step 2:
8a + 4b + 4 = 10.
4. Subtract 4 from both sides to isolate the terms with A and B:
8a + 4b = 6.
5. To find f'(-1), we need to differentiate the function f(x) with respect to x. Taking the derivative gives us:
f'(x) = 3ax^2 + 2bx + 3.
6. Substitute x = -1 into the derivative f'(-1) and set it equal to 14.
f'(-1) = 3a(-1)^2 + 2b(-1) + 3 = 3a - 2b + 3 = 14.
7. Simplify the equation obtained in step 6:
3a - 2b = 11.
Now we have a system of two linear equations:
8a + 4b = 6, (equation from step 4)
3a - 2b = 11. (equation from step 7)
We can solve this system of equations using various methods, such as substitution or elimination.
Let's solve it using the elimination method:
8a + 4b = 6 (multiply this equation by -3 to eliminate the a term)
3a - 2b = 11
-24a - 12b = -18
3a - 2b = 11
Now we can add both equations together and eliminate the b term:
(-24a - 12b) + (3a - 2b) = -18 + 11
-21a - 14b = -7
Simplifying further:
-21a - 14b = -7.
We can divide both sides of this equation by -7 to simplify it further:
3a + 2b = 1.
Now we have the following system of equations:
3a + 2b = 1 (equation from simplifying)
3a - 2b = 11 (equation from step 7)
Now, we can solve this system of equations using the elimination method again:
(3a + 2b) + (3a - 2b) = 1 + 11
6a = 12
Dividing both sides of the equation by 6, we obtain:
a = 2.
Substituting the value of a back into equation (1), we can solve for b:
3(2) + 2b = 1
6 + 2b = 1
2b = 1 - 6
2b = -5
b = -2.5.
Therefore, the values of A and B that satisfy the conditions f(2) = 10 and f'(-1) = 14 are A = 2 and B = -2.5.
f(x) = ax^3 + bx^2 + 3x - 2
f'(x) = 3ax^2 + 2bx + 3
f'(-1) = 3a - 2b + 3 = 14
3a - 2b = 11 **
also f(2) = 10
8a + 4b + 6 - 2 = 10
8a + 4b = 6
4a + 2b = 3 ***
add ** and ***
7a = 14
a = 2 , put that back into ** to find b