Max / min points of:
Y=5^(2x) * e^(3x^2)
To find the maximum and minimum points of the function Y = 5^(2x) * e^(3x^2), we need to take the derivative of the function and find the critical points.
Step 1: Take the derivative of Y with respect to x.
To differentiate Y with respect to x, we will use the product rule and the chain rule. The derivative of 5^(2x) * e^(3x^2) with respect to x is:
Y' = (5^(2x))' * e^(3x^2) + 5^(2x) * (e^(3x^2))'
Let's differentiate each term separately:
- (5^(2x))' = ln(5) * 2 * 5^(2x)
- (e^(3x^2))' = 2x * e^(3x^2) + e^(3x^2)
Now we substitute the derivatives back into Y':
Y' = ln(5) * 2 * 5^(2x) * e^(3x^2) + 5^(2x) * (2x * e^(3x^2) + e^(3x^2))
= 2 * ln(5) * 5^(2x) * e^(3x^2) + 5^(2x) * (2x + 1) * e^(3x^2)
Step 2: Find the critical points.
To find the critical points, we need to solve the equation Y' = 0. Let's set Y' = 0 and solve for x:
2 * ln(5) * 5^(2x) * e^(3x^2) + 5^(2x) * (2x + 1) * e^(3x^2) = 0
Since e^(3x^2) is always positive and non-zero, we can cancel it out:
2 * ln(5) * 5^(2x) + 5^(2x) * (2x + 1) = 0
Now we divide through by 5^(2x):
2 * ln(5) + 2x + 1 = 0
2x = -2 * ln(5) - 1
x = (-2 * ln(5) - 1) / 2
This gives us the value of x at which the function has a critical point.
Step 3: Determine whether it is a maximum or minimum point.
To determine if this critical point is a maximum or minimum, we need to evaluate the second derivative of Y.
Take the derivative of Y' with respect to x:
Y'' = [(2 * ln(5) * 5^(2x) * e^(3x^2))' + ((2x + 1) * e^(3x^2) * 5^(2x))'] * e^(-3x^2)
After differentiating and simplifying, we get:
Y'' = (4 * ln(5)^2 * 5^(2x) + 4 * ln(5) * (2x + 1) * 5^(2x) + (2x + 1)^2 * 5^(2x)) * e^(3x^2) - 6x^2 * e^(3x^2) * 5^(2x)
Now substitute the critical point x = (-2 * ln(5) - 1) / 2 back into Y'':
Y'' = (4 * ln(5)^2 * 5^((-2 * ln(5) - 1)/ 2) + 4 * ln(5) (-2 * ln(5) - 1 + 1) * 5^((-2 * ln(5) - 1)/2) + (-2 * ln(5) - 1)^2 * 5^((-2 * ln(5) - 1)/2)) * e^(3*((-2 * ln(5) - 1)/2)^2) - 6*((-2 * ln(5) - 1)/2)^2 * e^(3*((-2 * ln(5) - 1)/2)^2) * 5^((-2 * ln(5) - 1)/2)
Evaluate Y'' to determine whether it is positive or negative. If Y'' > 0, it is a minimum point. If Y'' < 0, it is a maximum point.
This process can be simplified if we use numerical methods or graphing software to find the critical points and analyze the second derivative.