Let f(x) be a differentiable function possessing an inverse function on f ^-1 (x). Suppose that the equation of the tangent line to the graph of f(x) at x=2 is y=3x-1. Then the equation of the tangent line to the graph of f^-1(x) at x=5 is...?
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y = 3x - 1 is tangent to graph of f at x = 2
f(2) = 5 (since this point is on line y = 3x - 1 which is tangent to f at x = 2)
f'(2) = 3 (slope of tangent line at x = 2)
Let g(x) = f⁻¹(x)
f(2) = 5 -----> g(5) = 2
g(f(x)) = x
Differentiate both sides with respect to x:
g'(f(x)) * f'(x) = 1
g'(f(x)) = 1/f'(x)
Let x = 2
g'(f(2)) = 1/f'(2)
g'(5) = 1/3
Tangent line passes through point (5,2) and has slope 1/3
y - 2 = 1/3 (x - 5)
y = 1/3 x - 5/3 + 2
y = 1/3 x + 1/3
f(2) = 5 (since this point is on line y = 3x - 1 which is tangent to f at x = 2)
f'(2) = 3 (slope of tangent line at x = 2)
Let g(x) = f⁻¹(x)
f(2) = 5 -----> g(5) = 2
g(f(x)) = x
Differentiate both sides with respect to x:
g'(f(x)) * f'(x) = 1
g'(f(x)) = 1/f'(x)
Let x = 2
g'(f(2)) = 1/f'(2)
g'(5) = 1/3
Tangent line passes through point (5,2) and has slope 1/3
y - 2 = 1/3 (x - 5)
y = 1/3 x - 5/3 + 2
y = 1/3 x + 1/3
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x=5--->y=2
the equation of the tangent line to the graph of f^-1(x) at x=5
y-2= (1/3)(x-5)
the equation of the tangent line to the graph of f^-1(x) at x=5
y-2= (1/3)(x-5)
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Even i can't solve calculus