How do I solve these derivative problems?

Question

Find tangent line at point (4,2) of the graph of f^-1 (inverse f(x)) if f(x)=x^3+2x-8?

Find the equation of the tangent line to the inverse at the given point

a) f(x)=x^3+7x+2 @(10,1)

b) f(x)= x^5+3x^3+7x+2 @(13,1)

Answer 1

f(x) = x³ + 2x - 8

The point (2,4) is on the graph of y = f(x), so (4,2) is on the graph of y = f^-1(x).

Slope of tangent line to the graph of f^-1(x) at (4,2) = [f^-1(4)]' = 1 / [f'(2)] = 1/14.

Equation of tangent line: y - 2 = (1/14)(x - 4)

y = (1/14)x + 12/7

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f(x) = x³ + 7x + 2

The point (1,10) is on the graph of y = f(x), so (10,1) is on the graph of y = f^-1(x).

Slope of tangent line to y = f^-1(x) at (10,1) = [f^-1(10)]' = 1 / [f'(1)] = 1/10

Equation of tangent line: y - 1 = (1/10)(x - 10)

y = (1/10)x

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Do the remaining problem in the same way.

Answer 2

I'll show you the solution of one problem - the rest is the same. If you need further assistance - please let me know

a) f(x)=x^3+7x+2 @(10,1)

the derivative of inverse function h is given by the following formula

h'(x) = 1 / f'( h(x) )

F' = 3x^2 + 7

h(10) = 1

So h'(10) = 1 / (3 * 1^2 + 7) = 1/10

As you know - the derivative is the slope of the tangent line

The equation of the line (in slope-point form): y - Y_a = m (x - X_a) at a point(X_a, Y_a) (10,1) slope m=1/10

So y-1=1/10(x-10)

Or y = 1/10 x