Find the slope and equation of the tangent line to the graph of the function at the given value of [tex]x[/tex].
[tex]f(x) = x^4 - 20x^2 + 64; \quad x = -1[/tex]
Answer (2)
The slope is the differential of the function.
Recall, if y = x^n, (dy/dx) = nx^(n-1).
y= x^4-20x^2+64; x = -1. To differentiate this, we do it for each term.
(dy/dx) = (4)(x^(4 -1)) - (2)(20x^(2-1) + 0*64x^(0-1) (Note 64 = 64x^0, x^0 =1) = (4)x^(3) - 40x^(1) + 0 = 4x^3 - 40x^1.
(dy/dx) = 4x^3 - 40x . Note at x = -1.
(dy/dx), x = -1, = 4(-1)^3 - 40(-1)
= -4 + 40 = 40 - 4 = 36.
Slope at x = -1 is 36.
Cheers.
The slope of the tangent line to the function f ( x ) = x 4 − 20 x 2 + 64 at x = − 1 is 36 . The equation of the tangent line is y = 36 x + 81 .
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