Given the function g (x) = 6x^3 + 54x^2 + 90x , find the first derivative, g'(x) .
Now, we want to know whether there is a local minimum or local maximum at x = − 1 , so we will use the second derivative test. Find the second derivative, g ' ' ( x ) . g ' ' ( x ) =
Evaluate g''(−1). g''(−1)=
I need the answer for (Evaluate g''(−1). g''(−1)=)
g' = 18x^2 + 108x + 90 = 18(x^2+6x+5)
so there are extrema at x=-1,-5
g" = 36(x+6)
extrema are
min if g" > 0 (g concave up)
max if g" < 0 (g concave down)
for g"(-1)
eh? You have the equation for g". Evaluate it at x = -1 !!
To find the second derivative of the function g(x) = 6x^3 + 54x^2 + 90x, we need to take the derivative of the first derivative. Let's start by finding the first derivative, g'(x):
g'(x) = 18x^2 + 108x + 90
Now, to find the second derivative, g''(x), we need to take the derivative of g'(x):
g''(x) = d/dx (18x^2 + 108x + 90)
= 36x + 108
To evaluate g''(−1), substitute x = -1 into the second derivative equation:
g''(−1) = 36(-1) + 108
= -36 + 108
= 72
Therefore, g''(−1) = 72.