Find the general formula for a line tangent to a given function at the point (x0,f(x0)).
The slope of the line must be f'(x0). Using the point-slope form of a line, y-y0=m(x-x0) y-f(x0)=f'(x0)(x-x0) y-f(x0)=f'(x0)x-f'(x0)x0 y=f'(x0)x-f'(x0)x0+f(x0) y=f'(x0)x+f(x0)-f'(x0)x0 y=(f'(x0))x+(f(x0)-f'(x0)x0) m=f'(x0), b=f(x0)-f'(x0)x0
Find the extremum of the parabola y=ax^2+bx+c without using calculus.
The extremum will occur at the vertex, x=-b/2a. Substitute this value into y=ax^2+bx+c. y=a(-b/2a)^2+b(-b/2a)+c=a(b^2/4a^2)+b(-b/2a)+c=b^2/4a-b^2/2a+c=b^2/4a-2b^2/4a+4ac/4a y= -(b^2-4ac)/4a
If an object is placed at position 1 m and starts moving with velocity 1 m/s, acceleration 1 m/s^2, jerk 1 m/s^3, all the way up to infinity, what will its position be after exactly one second, in meters?
In order for (d^n/dt^n)(x(t))=1 at t=0 for all non-negative integers n, we will require that x(t)=1+t+t^2/2!+t^3/3!+...=e^t. Thus its position (in meters) at t=1 is x(1)=e.