What is the instantaneous rate of change of f(x)=1x3−2x+5 at x=0?

This is just the value of the derivative at x = 0.differentiate using the chain ruleddx[f(g(x))]=f′(g(x))⋅g′(x)Rewrite 1x3−2x+5=(x3−2x+5)−1f(g(x))=(x3−2x+5)−1⇒f′(g(x))=−1(x3−2x+5)−2and g(x)=x3−2x+5⇒g′(x)=3x2−2now substitute these values into the derivative⇒f′(x)=−(x3−2x+5)−2×(3x2−2)=−(3x2−2)(x3−2x+5)2and f′(0)=−(−2)(5)2=225

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High School
Calculus and Analysis
Calculus 1
Derivatives

Ryan Farrell

Answered question

2022-03-16

What is the instantaneous rate of change of

f (x) = \frac{1}{x^{3} - 2 x + 5}

at x=0?

Answer & Explanation

Thonbrunnhng

Beginner2022-03-17Added 2 answers

This is just the value of the derivative at x = 0.
differentiate using the chain rule

\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)

Rewrite

\frac{1}{x^{3} - 2 x + 5} = {(x^{3} - 2 x + 5)}^{- 1}

f (g (x)) = {(x^{3} - 2 x + 5)}^{- 1}

\Rightarrow f^{'} (g (x)) = - 1 {(x^{3} - 2 x + 5)}^{- 2}

and

g (x) = x^{3} - 2 x + 5 \Rightarrow g^{'} (x) = 3 x^{2} - 2

now substitute these values into the derivative

\Rightarrow f^{'} (x) = - {(x^{3} - 2 x + 5)}^{- 2} \times (3 x^{2} - 2)

= \frac{- (3 x^{2} - 2)}{{(x^{3} - 2 x + 5)}^{2}}

and

f^{'} (0) = \frac{- (- 2)}{{(5)}^{2}} = \frac{2}{25}

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