Aiden Delacruz

2022-03-16

What is the instantaneous rate of change for $f\left(x\right)={x}^{3}-{x}^{2}-2x\cdot 4$ when x=2?

nida0694ii5

Beginner2022-03-17Added 6 answers

The instantaneous rate of change is the gradient of the slope at a given point, in this case, when x=2

The function is:$f\left(x\right)={x}^{3}-{x}^{2}-8x$

The gradient function is:${f}^{\prime}\left(x\right)=3{x}^{2}-2x-8$

Let x=2

therefore the gradient of the curve at this point is

f'(x)=3*4-2*2-8

f'(x)=12-4-8

f'(x)=0

The function is:

The gradient function is:

Let x=2

therefore the gradient of the curve at this point is

f'(x)=3*4-2*2-8

f'(x)=12-4-8

f'(x)=0

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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