Question

Use the limit definition of the derivative to find the instantaneous rate of change of

f(x)=5x^2+3x+3 at x=4

Answer 1

`f'(4) = 43`

Step-by-step explanation:

Given:
`f(x)=5x^2 + 3x + 3`

`\displaystyle f'(x)= \lim_(h \to 0) (f(x+h) - f(x))/(h)`

Note that

`f(x+h) = 5(x+h)^2 + 3(x+h) + 3`

`\:\:\;\:\:\:\:= 5(x^2 + 2hx + h^2) + 3x + 3h +3`

`\:\:\;\:\:\:\:= 5x^2 + 10hx + 5h^2 + 3x + 3h +3`

Substituting the above equation into the expression for f'(x), we can then write f'(x) as

`\displaystyle f'(x) = \lim_(h \to 0) (10hx + 3h + 5h^2)/(h)`

`\displaystyle\:\:\;\:\:\:\:= \lim_(h \to 0) (10x +3 +5h)`

`\:\:\;\:\:\:\:= 10x + 3`

Therefore,

`f'(4) = 10(4) + 3 = 43`