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2019-01-25, 14:05   #166
Dr Sardonicus

Feb 2017
Nowhere

115718 Posts

Quote:
 Originally Posted by jvang $$f(x) = x^3+2x+4$$, and $$g(x)$$ is its inverse. Find $$g(7)$$ without finding the formula for $$g(x)$$, then find $$g'(7)$$.
f(x) = x3 + 2*x + 4

y = x3 + 2*x + 4, point (1, 7)

y' = 3*x2 + 2 At (1, 7) y' = 5

Inverse function

x = y3 + 2*y + 4, point (7, 1)

1 = (3*y2 + 2)*y' At (7, 1) y' = 1/5

Using the "just transpose x and y" idea, the point-slope equation for the tangent line to y = f(x) at (1, 7) is

y - 7 = 5*(x - 1).

An equation for the tangent line to x = f(y) at (7, 1) is then

x - 7 = 5*(y - 1).

Casting this into point-slope form,

(1/5)*(x - 1) = y - 1.

Last fiddled with by Dr Sardonicus on 2019-01-25 at 14:11