Instantaneous Rate of Change/Derivative I am having a bit of trouble with a question on my Calc HW. Given P(x)= 3x^2+3, estimate the Instantaneous rate of change at x=5. This is what I have so far. f(x+h)=3(x+h)^2+3−(3x^2+3) =(3x+3h)(x+h)+3−(3x^2+3) =3x^2+3xh+3xh+3h^2+3−(3x^2+3) =3x^2+6xh+3h^2+3−3x^2−3 =(6xh+3h^2/h) =h(6x+3h)/h eliminate h and left with 6x+3h

Angeline Avila

Angeline Avila

Open question

2022-08-21

Instantaneous Rate of Change/Derivative
I am having a bit of trouble with a question on my Calc HW. Given P ( x ) = 3 x 2 + 3 , estimate the Instantaneous rate of change at x=5. This is what I have so far.
f ( x + h ) = 3 ( x + h ) 2 + 3 ( 3 x 2 + 3 )
= ( 3 x + 3 h ) ( x + h ) + 3 ( 3 x 2 + 3 )
= 3 x 2 + 3 x h + 3 x h + 3 h 2 + 3 ( 3 x 2 + 3 )
= 3 x 2 + 6 x h + 3 h 2 + 3 3 x 2 3
= ( 6 x h + 3 h 2 / h )
= h ( 6 x + 3 h ) / h
eliminate h and left with 6x+3h
I know this is incorrect, but don't know where I went wrong. I know for the last step, you have to plug in 5 to x and solve to get the Rate of change.

Answer & Explanation

doganomyt

doganomyt

Beginner2022-08-22Added 7 answers

6 x h + 3 h 2 h = 6 x + 3 h
Which approaches 6x as h approaches 0.
Plug in x=5:
6 × 5 = 30
Sorry to be so brief. I'm on a smartphone.
What you've got looks good, although it's difficult to read until it's formatted... I think that's in the works.
Make your first line
f ( x + h ) f ( h ) h

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