How do you find the instantaneous rate of change of the function x 3 + 2 x 2 + x when x=1?

Question

How do you find the instantaneous rate of change of the function       x          3        +  2      x          2        +  x when x=1?

Alisa Gilmore · Accepted Answer

Find the value of the derivative at x=1:To find the derivative of the function, use the power rule:      f    ′    (  x  )  =  3      x          2        +  4  x  +  1The instantaneous rate of change and slope of the tangent line at x=1 is:      f    ′    (  1  )  =  3  (  1      )          2        +  4  (  1  )  +  1  =  8

Yahir Tucker · Accepted Answer

Given:      x          3        +  2      x          2        +  x  ,  x  =  1      d          d      x        (      x          3        +  2      x          2        +  x  )  =  (  x  +  1  )  (  3  x  +  1  )Evaluate the derivative at x=1  (      d          d      x        (      x          3        +  2      x          2        +  x  )  )            |              (      x      =      1      )        =  (  (  x  +  1  )  (  3  x  +  1  )  )            |              (      x      =      1      )        =  8Therefore, the instantaneous rate of change of   f  (  x  )  =      x          3        +  2      x          2        +  x at x=1 is 8

How do you find the instantaneous rate of change of the function x <mrow class="MJX-TeXA

Answered question

Answer & Explanation

New Questions in Calculus 1