tangent calc problems!?

Question

that i have no idea how to get..

1. find the equation of the tangent line to y=f(x), x=a.
f(x) = x^3 + x, a=1

2. use the position function f(t) to find the velocity at time t=a.
f(x) = -16t^2 + 5, a=2

what do you mean by 'take the derivitive'?

Answer 1

1)
The line tangent to a point on f(x) has the same slope as f(x) at that point. To find the slope, take the derivative and put in your value for x:
f'(x) = 3x² + 1
f'(1) = 3 + 1 = 4
Now find the value of f(x) at x=1:
f(1) = 1 + 1 = 2
So the line tangent to f(x) at x=1 has a slope of 4 and passes through the point (x,f(x))=(1,2). You can use the point-slope form of a line:
(y - y1) = m(x - x1)
y - 2 = 4(x - 1)
y - 2 = 4x - 4
──► y = 4x -2

2)
The velocity of f(t) at a time t is the slope of f(t) at t.
Take the derivative of f(t) and plug in your value for t=a:
v(t) = f'(t) = -32t
v(2) = f'(2) = -64

Answer 2

To take the derivative means to compute the derivative of the function f at a given point. The derivative of f at x is defined as lim h -> 0 (f(x + h) - f(x))/h if such limit exists.

Blue gave you a very good answer. But frankly, if you don't know what a derivative is and has no idea what ths problem is about, then you should study harder. Asking people to solve your problems won't work.

Answer 3

1. (y-2)=4(x-1)
y=4x-2

2.f'(2)=-59