Answer to Question #333719 in Calculus for Zaib

Question

find 'f(t) by definition if f(t)=4t^2 + t also find tangent at t=2

Answer 1

a) Given $f(t)=4t^2+t.$

$f'(t)=\lim\limits_{h\to 0}\dfrac{4(t+h)^2+t+h-(4t^2+t)}{h}$

$=\lim\limits_{h\to 0}\dfrac{8th+4h^2+h}{h}$

$=\lim\limits_{h\to 0}(8t+4h+1)$

$=8t+0+1=8t+1$

$f'(t)=8t+1$

b)

Given $t=2.$

$f(2)=4(2)^2+2=18$

$f'(2)=8(2)+1=17$

$y-18=17(t-2)$

The equation of the tangent at $t=2$ in slope-intercept form is

$y=17t-16$