# find the values of a nad b if the function f(x)=2x^3 + ax^2 + bx + 36 has a local max when x=-4 and a min when x=5

First you calculate the derivative:

f'(x)=6x^2 + 2ax + b (1)

At the local maximum and minumum f' is zero. If a polynomial is zero at some point p, then it must contain a factor (x-p). So, since you know that f' is zero at both x =-4 and x = 5 you know that f' must be of the form:

f'(x) = A(x+4)(x-5) (2)

the two factors (x+4) and (x-5) make the right hand side a second degree polynomial, so A must be a constant. If f' were a third degree function then A would have been an unknown linea function. From (1) and (2) you find

6x^2 + 2ax + b = A (x+4)(x-5)

The coefficient of x^2 on the left hand side is 6, on the right hand side it is A, so you find that A = 6.

6 (x+4)(x-5) = 6 x^2 -6 x -120,

so 2a = -6 --> a = -3

and b = -120.

thanks!

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## To find the values of a and b for the given function, f(x) = 2x^3 + ax^2 + bx + 36, such that it has a local maximum at x = -4 and a minimum at x = 5, follow these steps:

Step 1: Calculate the derivative of the function, f'(x), with respect to x.

f'(x) = d/dx (2x^3 + ax^2 + bx + 36)

= 6x^2 + 2ax + b

Step 2: Set the derivative equal to zero and solve for x to find the critical points.

6x^2 + 2ax + b = 0

Step 3: Since the function has a local maximum at x = -4, substitute -4 for x in the above equation and solve for a and b.

6(-4)^2 + 2a(-4) + b = 0

96 - 8a + b = 0

Step 4: Similarly, substitute 5 for x in the equation from step 2 to find another equation relating a and b.

6(5)^2 + 2a(5) + b = 0

150 + 10a + b = 0

Step 5: Now you have two equations (from steps 3 and 4) with two variables (a and b). Solve this system of equations to find the values of a and b.

Using the method of substitution or elimination, solve the system of equations:

Equation 1: 96 - 8a + b = 0

Equation 2: 150 + 10a + b = 0

Solving these equations simultaneously will give the values of a and b.

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