Question

Show that the slope of the tangent line to the graph of f(x) at P is the reciprocal of the slope of the graph of g(x) at the conjugate point Q

f(x)=−2x³+1, point P(1,−1)
g(x)=∛(1-x)/2 conjugate point Q(−1,1)

Answer 1

The slope of the tangent line to the graph of f(x) at point P is the reciprocal of the slope of the graph of g(x) at the conjugate point Q. To show this, find the slopes of the two graphs at P and Q and demonstrate that they are reciprocals of each other.

Step-by-step explanation:

The slope of the tangent line to the graph of f(x) at point P is the reciprocal of the slope of the graph of g(x) at the conjugate point Q. To show this, we can find the slopes of the two graphs at P and Q and demonstrate that they are reciprocals of each other.

Find the slope of the tangent line to the graph of f(x) at point P:

To find the slope of the tangent line, we need to find the derivative of f(x) and evaluate it at x = 1. The derivative of f(x) = -2x³ + 1 is f'(x) = -6x². Evaluating f'(x) at x = 1, we get f'(1) = -6(1)² = -6.

Find the slope of the graph of g(x) at the conjugate point Q:

To find the slope of the graph of g(x), we need to find the derivative of g(x) and evaluate it at x = -1. The derivative of g(x) = ∛(1-x)/2 is g'(x) = -1/(6∛(x-1)²). Evaluating g'(x) at x = -1, we get g'(-1) = -1/(6∛((-1)-1)²) = -1/6.

Show that the slopes are reciprocal:

The slope of the tangent line to the graph of f(x) at point P is f'(1) = -6, and the slope of the graph of g(x) at the conjugate point Q is g'(-1) = -1/6. The reciprocal of -6 is -1/(-6) = 1/6, which is equal to -1/6.

Therefore, the slope of the tangent line to the graph of f(x) at point P is the reciprocal of the slope of the graph of g(x) at the conjugate point Q.