How to find instantaneous rate of change for $f(x)={x}^{2}+3x+4$ when x=4?

oleifere45
2022-06-25
Answered

How to find instantaneous rate of change for $f(x)={x}^{2}+3x+4$ when x=4?

You can still ask an expert for help

Elianna Douglas

Answered 2022-06-26
Author has **23** answers

Explanation:

This is just the value of f'(x) at x = 4

f'(x) = 2x + 3

and f'(4) = 2(4) + 3 = 8 + 3 = 11

This is just the value of f'(x) at x = 4

f'(x) = 2x + 3

and f'(4) = 2(4) + 3 = 8 + 3 = 11

Kapalci

Answered 2022-06-27
Author has **9** answers

Given:

${x}^{2}+3x+4,x=4$

$\frac{d}{dx}({x}^{2}+3x+4)=2x+3$

x=4

$(\frac{d}{dx}({x}^{2}+3x+4)){|}_{(x=4)}=(2x+3){|}_{(x=4)}=11$

$f(x)={x}^{2}+3x+4$ at x=4 is 1

${x}^{2}+3x+4,x=4$

$\frac{d}{dx}({x}^{2}+3x+4)=2x+3$

x=4

$(\frac{d}{dx}({x}^{2}+3x+4)){|}_{(x=4)}=(2x+3){|}_{(x=4)}=11$

$f(x)={x}^{2}+3x+4$ at x=4 is 1

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