consider the following function. g(x)=4x^4 - 8x^3 + 6 find the first and second derivatives of the function…

consider the following function. g(x)=4x^4 - 8x^3 + 6 find the first and second derivatives of the function. g(x)= g(x)= find the inflection point(s), if any, of the function. (if an answer does not exist, enter dne.) smaller x - value (x,y)= larger x - value (x,y)= need help? read it

consider the following function. g(x)=4x^4 - 8x^3 + 6 find the first and second derivatives of the function. g(x)= g(x)= find the inflection point(s), if any, of the function. (if an answer does not exist, enter dne.) smaller x - value (x,y)= larger x - value (x,y)= need help? read it

Answer

Explanation:

Step1: Find the first - derivative

Use the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$. For $g(x)=4x^{4}-8x^{3}+6$, we have $g'(x)=4\times4x^{3}-8\times3x^{2}=16x^{3}-24x^{2}$.

Step2: Find the second - derivative

Differentiate $g'(x)$ using the power rule. $g''(x)=16\times3x^{2}-24\times2x = 48x^{2}-48x$.

Step3: Find the inflection points

Set $g''(x) = 0$. So, $48x^{2}-48x=48x(x - 1)=0$. Solving $48x(x - 1)=0$ gives $x = 0$ and $x = 1$. To find the $y$ - values, substitute $x$ into the original function $g(x)$. When $x = 0$, $g(0)=4\times0^{4}-8\times0^{3}+6=6$. When $x = 1$, $g(1)=4\times1^{4}-8\times1^{3}+6=4 - 8 + 6=2$.

Answer:

$g'(x)=16x^{3}-24x^{2}$ $g''(x)=48x^{2}-48x$ smaller $x$ - value $(x,y)=(0,6)$ larger $x$ - value $(x,y)=(1,2)$