What is the instantaneous rate of change of f(x)=1x2−x+3 at x=0?

Find f'(x) or the instantaneous rate of change of f(x) at x.=−1(x2−x+3)2⋅ddx(x2−x+3)(power rule: ddx(xn)=nxn−1 and chain rule: ddx(f(g(x)))=f′(g(x))g′(x))=−1(x2−x+3)2⋅(2x−1)plug in 0 for x:f′(0)=−1(02−0+3)2⋅(2(0)−1)f′(0)=−132⋅(−1)f′(0)=19

Explanation:the instantaneous rate of change of f(x) at x=0 is f'(0)differentiate using the chain rulegiven f(x)=g(h(x)) thenf′(x)=g′(h(x))×(x2−x+3)−1⇒f′(x)=−(x2−x+3)−2×ddx(x2−x+3)=−2x−1(x2−x+3)2⇒f′(0)=−−132=19

Find f'(x) or the instantaneous rate of change of f(x) at x.=−1(x2−x+3)2⋅ddx(x2−x+3)(power rule: ddx(xn)=nxn−1 and chain rule: ddx(f(g(x)))=f′(g(x))g′(x))=−1(x2−x+3)2⋅(2x−1)plug in 0 for x:f′(0)=−1(02−0+3)2⋅(2(0)−1)f′(0)=−132⋅(−1)f′(0)=19

Explanation:the instantaneous rate of change of f(x) at x=0 is f'(0)differentiate using the chain rulegiven f(x)=g(h(x)) thenf′(x)=g′(h(x))×(x2−x+3)−1⇒f′(x)=−(x2−x+3)−2×ddx(x2−x+3)=−2x−1(x2−x+3)2⇒f′(0)=−−132=19

Check the solution to "What is the instantaneous rate of change of..."

pevetaty

Answered question

2022-03-15

What is the instantaneous rate of change of

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

at x=0?

Answer & Explanation

Raster02f

Beginner2022-03-16Added 1 answers

Find f'(x) or the instantaneous rate of change of f(x) at x.

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

(power rule:

\frac{d}{d x} (x^{n}) = n x^{n - 1}

and chain rule:

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

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

plug in 0 for x:

f^{'} (0) = - \frac{1}{{(0^{2} - 0 + 3)}^{2}} \cdot (2 (0) - 1)

f^{'} (0) = - \frac{1}{3^{2}} \cdot (- 1)

f^{'} (0) = \frac{1}{9}

Ciara Hoffman

Beginner2022-03-17Added 5 answers

Explanation:
the instantaneous rate of change of f(x) at x=0 is f'(0)
differentiate using the chain rule
given f(x)=g(h(x)) then

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

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

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

\Rightarrow f^{'} (0) = - \frac{- 1}{3^{2}} = \frac{1}{9}

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 1

Didn't find what you were looking for?