How do you find the instantaneous rate of change of the function #y=4x^3+2x-3# when x=2?

Question

Isabella Ackerman · Accepted Answer

#50# #"the instantaneous rate of change at a point is the"# #"value of the derivative at that point, that is it is a"# #"measure of the slope of the tangent line"# #dy/dx=12x^2+2# #x=2tody/dx=12(2)^2+2=50#

Ezra Adams · Answer

undefined To find the instantaneous rate of change of the function $ y = 4x^3 + 2x - 3 $ when $ x = 2 $, you can use the derivative of the function. The derivative of the function gives the slope of the tangent line to the function at any given point. So, you need to find the derivative of the function with respect to $ x $, and then evaluate it at $ x = 2 $.

First, find the derivative of the function:

$$ \frac{dy}{dx} = 12x^2 + 2 $$

Next, evaluate the derivative at $ x = 2 $:

$$ \frac{dy}{dx} \bigg|_{x=2} = 12(2)^2 + 2 $$

$$ = 12(4) + 2 $$

$$ = 48 + 2 $$

$$ = 50 $$

So, the instantaneous rate of change of the function $ y = 4x^3 + 2x - 3 $ when $ x = 2 $ is $ 50 $.

Isabella Ackerman · Answer

undefined To find the instantaneous rate of change of the function y = 4x^3 + 2x - 3 when x = 2, we need to find the derivative of the function with respect to x, and then evaluate it at x = 2.

The derivative of the function y = 4x^3 + 2x - 3 with respect to x is given by:

dy/dx = 12x^2 + 2.

Now, we evaluate the derivative at x = 2:

dy/dx = 12(2)^2 + 2
      = 12(4) + 2
      = 48 + 2
      = 50.

So, the instantaneous rate of change of the function y = 4x^3 + 2x - 3 when x = 2 is 50.