1) solve:e^x(x^2-4)=0
e^x=0 and then (x+2)(x-2) so x=2, -2
2) differentiate: y=ln(6x^2 - 3x + 1)
1/(6x^2 - 3x + 1) * 12x-3
3) differentiate: y=e^-3x+2
-3 * e^-3x+2
4) evaluate: 2^4-x=8
2^4-x = 2^3
4-x = 3
-x=-1 so x=1
5) differentiate: x^3 + y^3 -6 =0
3x^2 + 3y^2
6) A rectangular garden has an area of 100 square meters for which the amount of fencing needed to surround the garden should be as small as possible.
a) draw a picture of a rectangle and select appropriate letters for the dimensions
I chose x and y
b)determine the objective and constraint equations
objective: A=xy
constraint= 100=xy
c) find the optimal values for the dimensions.
I am not sure for this one
Thank you for your help!!!
1. e^x can never equal zero, but can approach it as x>>-infinity
But the roots are in fact -2,2
2 correct
3 e^(-3x+2)=e^-3x * e^2
then y'= -3e^-3x * e^2=-3e^(-3x+2)
4, correct
5. No.
x^3+y^3=0 You can do implicit differentation, but I am not certain you know that.
y^3=x^3
y= cubroot x^3=x
y'=1
6) xy=100 constraint
Perimeter= 2x+2y objective
dP/dx= 2+ d2(200/x)/dx= 2-400/x^2 =0
2x^2=400
x=10 Y=10 is minimum fencing
5) (x^3+y^3-6)'=3x^2+3y^2*y'=0
y'=-x^2/y^2, where y=(6-x^3)^(1/3)
I agree, Mgraph, thanks.
Thank you!!
1) To solve the equation e^x(x^2-4)=0, first, we set e^x=0 and solve for x. However, e^x cannot be equal to 0 since e^x is always positive for any real value of x.
Next, we set the expression inside the parenthesis equal to 0: (x^2-4)=0.
This is a quadratic equation in the form of (x-a)(x+b)=0, where a=2 and b=-2. Therefore, by setting each factor to 0, we get x=2 and x=-2 as the solutions to the quadratic equation.
2) To differentiate the function y=ln(6x^2 - 3x + 1), we apply the chain rule of differentiation. The chain rule states that if y=f(g(x)), then dy/dx=f'(g(x))*g'(x).
In this case, f(u) = ln(u) and u = 6x^2 - 3x + 1. Therefore, we have y = f(u), and we need to find dy/dx.
Using the chain rule, the derivative of y with respect to x is given by: dy/dx = f'(u) * u', where f'(u) is the derivative of ln(u) and u' is the derivative of u with respect to x.
The derivative of ln(u) is 1/u, and the derivative of u = 6x^2 - 3x + 1 with respect to x is 12x - 3.
So, dy/dx = (1/u) * (12x - 3) = (12x - 3)/(6x^2 - 3x + 1).
3) To differentiate the function y=e^-3x+2, we apply the chain rule again.
In this case, f(u) = e^u and u = -3x + 2. Therefore, we have y = f(u), and we need to find dy/dx.
Using the chain rule, the derivative of y with respect to x is given by: dy/dx = f'(u) * u'.
The derivative of e^u is e^u, and the derivative of u = -3x + 2 with respect to x is -3.
So, dy/dx = e^u * -3 = -3 * e^-3x + 2.
4) To evaluate the equation 2^(4-x) = 8, we use the property of exponentiation that says if a^b = c, then b = log_a(c).
In this case, we have 2^(4-x) = 8. To simplify, we can rewrite 8 as 2^3.
So, we have 2^(4-x) = 2^3.
According to the property mentioned above, we equate the exponents:
4 - x = 3.
Solving this equation for x, we get -x = -1, which implies x = 1.
5) To differentiate the equation x^3 + y^3 - 6 = 0, we take the derivative of each term with respect to x.
The derivative of x^3 is 3x^2 (using the power rule).
The derivative of y^3 is 3y^2 * dy/dx (using the chain rule, as we differentiate with respect to x).
However, since the equation relates x and y, we need to find dy/dx in terms of x and y. This can be done by taking the derivative of the constraint equation x^3 + y^3 - 6 = 0 and solving for dy/dx.
Differentiating both sides of the equation gives us:
3x^2 + 3y^2 * dy/dx = 0.
Solving for dy/dx, we find dy/dx = -3x^2 / (3y^2) = -x^2 / y^2.
So, the derivative of x^3 + y^3 - 6 = 0 is 3x^2 - x^2 / y^2.
6) For the rectangular garden, we want to find the dimensions that will minimize the amount of fencing needed to surround the garden.
a) To represent the dimensions, let's choose x as the length of the garden and y as the width of the garden. Thus, we have a rectangle with sides x and y.
b) The objective equation is the equation that represents what we want to optimize. In this case, we want to minimize the amount of fencing needed, which is determined by the perimeter of the rectangle. The perimeter is given by P = 2x + 2y.
The constraint equation is the equation that represents the condition to be satisfied. Here, we are given that the area of the garden is 100 square meters, so the constraint equation is A = xy = 100.
c) To find the optimal values for the dimensions, we need to solve the objective equation and the constraint equation simultaneously. This can be done using optimization techniques such as Lagrange multipliers or substitution.
In this case, to find the dimensions that minimize the amount of fencing, we can use substitution to solve for one variable in terms of the other from the constraint equation and substitute it into the objective equation.
Here, we can solve the constraint equation for y: y = 100/x. Substituting this value of y into the objective equation, we get P = 2x + 2(100/x).
Differentiate P with respect to x and set the derivative equal to zero to find the critical points. Then, find the values of x and y that correspond to those critical points.
Finally, substitute the found values of x and y into the objective equation P = 2x + 2y to find the minimum amount of fencing needed.
Unfortunately, without specific numerical values, we cannot proceed further to find the optimal values for the dimensions.