Let f(x) = 1/x^2 (a) use the definition of the derivatve to find f'(x). (b) find the equation of the tangent line at x=2

(a)

(b)

Step-by-step explanation:

The given function is

                 .... (1)

According to the first principle of the derivative,

Cancel out common factors.

By applying limit, we get

                        .... (2)

Therefore .

(b)

Put x=2, to find the y-coordinate of point of tangency.

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

Substitute x=2 in equation 2.

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

greater than one

step-by-step explanation: