Question

Given the following function, what is the instantaneous rate of change at X=2?y=1/3x^3+2x^2-4x+1

Answer 1

`\text{8 (option B)}`Step-by-step explanation:

`\begin{gathered} \text{Given function:} \\ y\text{ = }(1)/(3)x^3+2x^2-4x+1​ \end{gathered}`

To get the instantaneous rate of change, first we need to differentiate the function with respect to x:

`\begin{gathered} (dy)/(dx)\text{ = }(1)/(3)(3x^(3-1))+2(2x^(2-1))-4(1x^(1-1))\text{ + 0} \\ \frac{\mathrm{d}y}{dx}\text{ = }(1)/(3)(3x^2)+2(2x^1)-4(1x^0)\text{ + 0} \\ \frac{\mathrm{d}y}{dx}\text{ = }x^2+2(2x^{})-4(1)\text{ + 0} \\ \frac{\mathrm{d}y}{dx}\text{ = }x^2+4x-4\text{ + 0} \\ \frac{\mathrm{d}y}{dx}\text{ = }x^2+4x-4 \end{gathered}`

At x = 2

`\begin{gathered} \text{After differentiation, we will substitute 2 for x in the derivative:} \\ (dy)/(dx)=(2)^2\text{ + 4(2) - 4} \\ \frac{\mathrm{d}y}{dx}=4\text{ + 8 - 4} \\ \frac{\mathrm{d}y}{dx}=\text{ 8 (option B)} \end{gathered}`