For #f(x) =-x^-3+x^-2#, what is the equation of the line tangent to #x =2 #?

Question

Evelyn Agard · Accepted Answer

Here's two ways to do it. I get

#f_T(2) = 7/16 x - 3/4#

graph{(y - (1/4 - 1/16x))(y - (-x^(-3) + x^(-2))) = 0 [-1.127, 5.032, -1.836, 1.243]}

LINEARIZATION METHOD

The easy way to do it is via this equation:

#bb(f_T(a) = f(a) + f"'"(a)(x+a))#

where #f_T# is the function for the tangent line, #f# is the function, and #f'# is its derivative. #a# is the #x# value at which you are evaluating #f_T#.

When you take the derivative of polynomials, use the power rule:

#d/(dx)[x^n] = nx^(n-1)#

Therefore:

#f'(x) = -(-3)x^(-4) + (-2)x^(-3)#

#= 3/(x^4) - 2/(x^3)#

And so at #a = 2#,

#color(green)(f'(2)) = 3/(2^4) - 2/(2^3)#

#= 3/16 - 2/8#

#= 3/16 - 4/16 = color(green)(-1/16)#

And finally,

#color(green)(f(2)) = -(2)^(-3) + (2)^(-2)#

#= -1/8 + 1/4#

#= -1/8 + 2/8 = color(green)(1/8)#

So, using the first equation, we then get:

#color(blue)(f_T(2)) = f(2) + f'(2)(x+2)#

#= 1/8 - 1/16(x+2)#

#= 2/16 - 1/16 x + 2/16#

#= color(blue)(-1/16 x + 1/4)#

REGULAR WAY

Or, you could take the derivative, evaluate it at #x = 2#, then extrapolate to the #y#-intercept. It's a bit harder but does not require that you memorize much of an equation, other than slope is #(Deltay)/(Deltax)#.

#color(green)(f'(2) = -1/16)#, from before.

So, you have the slope. Your equation is so far:

#f_T(2) = -1/16x + ???#

Then, plug in #f(2)# and use the slope formula to find the #y#-intercept.

#color(green)(f(2) = 1/8)#, from before.

Therefore:

#(Deltay)/(Deltax) = (1/8 - y_1)/(2 - 0) = -1/16#

Now, solving for #y_1#:

#-1/8 = 1/8 - y_1#

#color(green)(y_1) = 1/8 + 1/8 = color(green)(1/4)#

So the end result is the same:

#color(blue)(f_T(2) = -1/16 x + 1/4)#

This is just more conceptual/visual.

Cameron Alexander · Answer

undefined The equation of the line tangent to f(x) at x = 2 is y = -4x + 9.