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

$f'(4) = 43$

Explanation:

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

$\displaystyle f'(x)= \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$

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

$\displaystyle f'(x) = \lim_{h \to 0} \dfrac{10hx + 3h + 5h^2}{h}$

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

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

Therefore,

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