Please check, if there is something wrong please explain what I did wrong. Thank you!
Calculate the d^2y/dx^2.
y= e^-x + e^x
y' = e^x - e^-x
y'' = e^x + e^-x
Find the x-coordinace of all critical points of the given function. determine whether each critical point is a relative minimum, maximum, or neither by first applying the second derivative test, and, if the test fails by some other method.
g(x) = 2x^3 - 6x + 3
g'(x) = 6x^2 -6
g"(x) = 12x
max = -1
min = 1
critical = -1,1
Looks good to me.
Nice work.
To calculate the second derivative of y, which is denoted as d^2y/dx^2, you correctly first found the first derivative of y, which is denoted as dy/dx.
To find the second derivative, we need to differentiate the first derivative with respect to x once again. In this case, the first derivative is y' = e^x - e^-x.
Now, differentiating y' with respect to x, we follow the power rule for differentiation, meaning we multiply each term by its power and decrease the power by 1. Applying this rule, we get:
(d/dx)(e^x - e^-x) = e^x + e^-x
Therefore, the second derivative of y, which is denoted as d^2y/dx^2, is y'' = e^x + e^-x.
Moving on to the next part of your question, we have the function g(x) = 2x^3 - 6x + 3. To find the critical points of this function, we need to find the values of x where the first derivative g'(x) equals zero or doesn't exist.
You correctly calculated the first derivative g'(x) = 6x^2 - 6. Now, we set this equal to zero and solve for x:
6x^2 - 6 = 0
6x^2 = 6
x^2 = 1
x = ±1
So, we have two critical points: x = -1 and x = 1.
To determine whether each of these critical points is a relative minimum, maximum, or neither, we can use the second derivative test. The second derivative g''(x) = 12x tells us about the concavity of the function.
At x = -1:
g''(-1) = 12(-1) = -12, which is negative. According to the second derivative test, when the second derivative is negative, the function is concave down. Thus, x = -1 is a relative maximum.
At x = 1:
g''(1) = 12(1) = 12, which is positive. According to the second derivative test, when the second derivative is positive, the function is concave up. Therefore, x = 1 is a relative minimum.
In summary, the critical points of g(x) = 2x^3 - 6x + 3 are x = -1 and x = 1. x = -1 is a relative maximum, and x = 1 is a relative minimum.