What is the instantaneous rate of change of #f(x)=1/(x^3-2x+5 )# at #x=1 #?

Question

What is the instantaneous rate of change of #f(x)=1/(x^3-2x+5      )# at #x=1 #?

Evelyn Agard · Accepted Answer

# -1/16#

This represents the derivative's value at x = 1.

rewrite f(x) as f(x) = # (x^3 - 2x + 5)^-1 #

now differentiate using the #color(blue)" chain rule " #

#d/dx[f(g(x)) ] = f'(g(x)). g'(x) # #"-----------------------------------------------------"#

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

#rArr f'(g(x)) = -(x^3-2x+5)^-2 #

and g(x) # = x^3-2x+5 rArr g'(x) = 3x^2 - 2 # #"--------------------------------------------------------"#

#rArr f'(x) = -(x^3-2x+5)^-2 .(3x^2-2) #

# = (-(3x^2-2))/(x^3-2x+5)^2 #

#rArr f'(1) = (-(3-2))/(1-2+5)^2 = (-1)/(4)^2 = -1/16 #

Luke Alvarez · Answer

undefined To find the instantaneous rate of change of the function $ f(x) = \frac{1}{x^3 - 2x + 5} $ at $ x = 1 $, we use the concept of the derivative. The derivative of a function at a specific point represents its instantaneous rate of change at that point.

To find the derivative of $ f(x) $, we'll use the quotient rule:

$$ \frac{d}{dx}\left(\frac{1}{x^3 - 2x + 5}\right) = \frac{-3x^2 - 2}{(x^3 - 2x + 5)^2} $$

Now, to find the instantaneous rate of change at $ x = 1 $, we substitute $ x = 1 $ into the derivative expression:

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

Solving this expression gives us the instantaneous rate of change of $ f(x) $ at $ x = 1 $.