Find the range of values of X for which the function is increasing
Y=8X+X^2-X^3/3
To find the range of values of X for which the function is increasing, we need to find the derivative of the function and determine when the derivative is positive.
The derivative of the function Y = 8X + X^2 - (X^3/3) is:
Y' = 8 + 2X - X^2
To find when the derivative is positive, we set Y' > 0 and solve for X:
8 + 2X - X^2 > 0
Rearranging the inequality:
X^2 - 2X - 8 < 0
Factoring, we get:
(X - 4)(X + 2) < 0
This inequality represents when the function is increasing. To solve for X, we set each factor equal to 0 and find the sign changes:
X - 4 = 0, X = 4 (sign change from negative to positive)
X + 2 = 0, X = -2 (sign change from positive to negative)
Plotting these points on a number line:
-∞ -2 4 ∞
Since the inequality is less than 0, the function is increasing when X is between -2 and 4.
Therefore, the range of values of X for which the function is increasing is -2 < X < 4.
To find the range of values of X for which the function is increasing, we need to determine the critical points of the function by finding its derivative and setting it equal to zero.
Step 1: Find the derivative of the function:
Y' = 8 + 2X - X^2
Step 2: Set the derivative equal to zero and solve for X to find the critical points:
8 + 2X - X^2 = 0
Rearranging the equation, we have:
X^2 - 2X - 8 = 0
Step 3: Solve the quadratic equation:
Factoring the quadratic equation, we have:
(X - 4)(X + 2) = 0
Setting each factor equal to zero, we find two critical points:
X - 4 = 0 => X = 4
X + 2 = 0 => X = -2
Step 4: Determine the intervals where the function is increasing or decreasing using the critical points:
To do this, we can use a number line and test values in the derivative of the function.
Testing a value less than -2, let's say -3:
Y' = 8 + 2(-3) - (-3)^2 = 8 - 6 - 9 = -7
Since the derivative is negative, the function is decreasing to the left of -2.
Testing a value between -2 and 4, let's say 0:
Y' = 8 + 2(0) - 0^2 = 8
Since the derivative is positive, the function is increasing between -2 and 4.
Testing a value greater than 4, let's say 5:
Y' = 8 + 2(5) - 5^2 = 8 + 10 - 25 = -7
Since the derivative is negative, the function is decreasing to the right of 4.
Step 5: Write the range of values of X for which the function is increasing:
The function is increasing for values of X between -2 and 4, inclusive.
Therefore, the range of values of X for which the function is increasing is (-2, 4].
To determine the range of values for which the function is increasing, we need to find the first derivative of the function and locate the intervals where it is positive.
First, let's find the derivative of the function Y=8X+X^2-X^3/3 with respect to X.
The derivative of Y = 8X + X^2 - X^3/3 can be found using the power rule and the constant multiple rule of differentiation:
Y' = d/dX (8X) + d/dX (X^2) - d/dX (X^3/3)
= 8 + 2X - d/dX (X^3/3)
To find the derivative of X^3/3, we apply the power rule and divide the exponent by the original exponent:
d/dX (X^3/3) = (1/3)*d/dX (X^3)
= (1/3) * 3X^2
= X^2
Now we can substitute this back into the equation:
Y' = 8 + 2X - X^2
To determine the intervals where the function is increasing, we need to find the values of X for which Y' > 0.
Setting Y' > 0, we have:
8 + 2X - X^2 > 0
Rearranging the equation:
X^2 - 2X - 8 < 0
Now we need to solve this inequality to find the range of values for X. We can start by factoring the quadratic equation:
(X - 4)(X + 2) < 0
Now we have two factors: (X - 4) and (X + 2). We need to determine the sign of each factor for specific values of X to determine the intervals where the inequality is satisfied.
Setting each factor to zero and solving for X:
X - 4 = 0 --> X = 4
X + 2 = 0 --> X = -2
Now we can create a sign chart:
-∞ -2 4 +∞
-----------------------
+ - + -
We can see from the sign chart that the inequality (X - 4)(X + 2) < 0 is satisfied when -2 < X < 4.
Therefore, the function Y=8X+X^2-X^3/3 is increasing for the range of values -2 < X < 4.