Solved: "What is the instantaneous rate of change of..."

High School
Calculus and Analysis
Calculus 1
Derivatives

melalcolicaoh3

Answered question

2022-03-16

What is the instantaneous rate of change of

f (x) = (x^{2} - 2) e^{x + 2}

at x=1?

Answer & Explanation

diesel817637dsf

Beginner2022-03-17Added 13 answers

The instantaneous rate of change is the same as the derivative of the function.
We need the product rule to differentiate this function and using it gives

f^{'} (x) = (x^{2} - 2) e^{x + 2} + 2 x e^{x + 2}

∴ f^{'} (1) = (1 - 2) e^{3} + 2 e^{3}

= e^{3}

Answer:

f^{'} (1) = e^{3}

elxanaijm

Beginner2022-03-18Added 5 answers

Given:

(x^{2} - 2) e^{x + 2}, x = 1

Solution:

\frac{d}{d x} ((x^{2} - 2) e^{x + 2}) = (x^{2} + 2 x - 2) e^{x + 2}

Derivative at x=1.

(\frac{d}{d x} ((x^{2} - 2) e^{x + 2})) {|_{{(x = 1)}} = ((x^{2} + 2 x - 2) e^{x + 2}) |}_{(x = 1)} = e^{3}

Answer:

f (x) = (x^{2} - 2) e^{x + 2}

at x=1 is

e^{3}

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