Question

Given the function g(x) = 6x ^ 3 + 45x ^ 2 + 72x find the first derivative , g^ prime (x); g^ prime (x)=

Answer 1

The Solution:

Given:

The first derivative gives:

$\begin{gathered} g^{\prime}(x)=3\times6x^2+(2\times45)x+72 \\ \\ g^{\prime}(x)=18x^2+90x+72 \end{gathered}$

So,

$g^{\prime}(-1)=18(-1)^2+90(-1)+72=18+72-90=0$

The second derivative gives:

$g^{\prime}^{\prime}(x)=36x+90$$g^{\prime}^{\prime}(-1)=36(-1)+90=-36+90=54$

By the second derivative test:

$\begin{gathered} g^{\prime}^{\prime}(-1)>0,\text{ then there is a local minimum at }x=-1. \\ \\ g^{\prime\prime}(-1)<0,\text{ then there is a local maximum at }x=-1. \end{gathered}$

Thus, the function g(x) has a local minimum at x= -1 since the second derivative is greater than zero at x = -1.

At x =-1, the graph og g(x) is concave up.