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

$$y=17t-16$$