Find the correct values for the equation -x^3 +bx^2 +cx + d, using this information, local min x= -5, local max (1,11). Point of inflection = -2.
I assume you mean
y or f(x) = -x^3 +bx^2 +cx + d
if so
y' = -3 x^2 + 2 b x + c
and
y" = -6 x +2 b
at x = -5 , y' = 0
-75 -10 b + c = 0
at x = 1 , y' = 0
-3 + 2 b + c = 0
so
2 b + c = 3
-10 b + c = 75
-----------------subtract
12 b = -72
b = -6
then
-12 + c = 3
c = -15
put in y = 11 at x = 1 to get d
11 = -1 -6 -15 + d
d = 33
Hmmm. I get
y = -x^3+bx^2+cx+d
y' = -3x^2+2bx+c
y" = -6x+2b
inflection at x = -2 means
-6x+2b = 0
b = -6
y = -x^3 - 6x^2 + cx + d
extrema at x = -5,1
y' = (x+5)(x-1)
= x^2+4x-5
we need -3x^2, so
y' = -3x^2-12x+15
so c = 15
y = -x^3 - 6x^2 + 15x + d
f(2) = 11, so
11 = -1-6+15+d
d = 3
y = -x^3 - 6x^2 + 15x + 3
then
-12 + c = 3
c = -15 No, +15 !!!!
put in y = 11 at x = 1 to get d
11 = -1 -6 -15 + d
d = 33
To find the correct values for the equation -x^3 + bx^2 + cx + d using the given information, we need to determine the values of b, c, and d.
1. Local minimum x = -5:
For a local minimum at x = -5, we know that the derivative of the function is zero at this point. Let's differentiate the function to find this point:
f'(x) = -3x^2 + 2bx + c
Setting f'(-5) = 0, we have:
-3(-5)^2 + 2b(-5) + c = 0
Simplifying:
75 - 10b + c = 0 (Equation 1)
2. Local maximum at (1,11):
For a local maximum at x = 1 and y = 11, we have two conditions:
a) The derivative of the function is zero at x = 1.
b) The function value at x = 1 is 11.
Let's differentiate the function as we did before:
f'(x) = -3x^2 + 2bx + c
Setting f'(1) = 0, we have:
-3(1)^2 + 2b(1) + c = 0
-3 + 2b + c = 0 (Equation 2)
Setting f(1) = 11, we have:
-(1)^3 + b(1)^2 + c(1) + d = 11
-b + c + d = 12 (Equation 3)
3. Point of inflection = -2:
For a point of inflection at x = -2, we know that the second derivative of the function is zero at this point. Let's find the second derivative:
f''(x) = -6x + 2b
Setting f''(-2) = 0, we have:
-6(-2) + 2b = 0
12 + 2b = 0
2b = -12
b = -6 (Equation 4)
Now, we have four equations (Equations 1, 2, 3, and 4) with the variables b, c, and d. We can solve these equations using substitution or elimination method to find the values of b, c, and d.