Question

the function f(x)=5x^2+3e^2x is invertible. give the slope of the normal line to the graph of inverse of f at x=3

Answer 1

because \(f\) is invertible by the inverse function theorem we have \(df^{-1}/dx=1/(df/dx)\)

Answer 2

since \(df/dx=10x+6e^{2x}\) and at \(x=3\) this yields \(10(3)+6e^{2(3)}=30+6e^6\) so the slope of the tangent to our inverse function is given by \(1/(30+6e^6)\). can you find the slope of a line perpendicular to our tangent function? if you guessed \(-(30+6e^6)\) you are correct