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Mathematics 14 Online
OpenStudy (anonymous):

Find the inverse function of f. f(x) = x^2 + 7x, x ≥ −7/2

OpenStudy (anonymous):

Put f(x)=y and using algebra find x in terms of y. But while taking square root, you need to consider the sign properly given the domain of the function is x>=-7/2

OpenStudy (anonymous):

so I'm coming up with \[\sqrt{x/7}\] but idk what I'm doing wrong. @anurag.luvs.india

OpenStudy (anonymous):

\[x^2+7x-y=0\] This is the equation you have in your hand. Taking y as dummy variable, solve for quadratic equation in terms of y using the formula. From there you get x in terms of y as\[x=(-7\pm(49+2y)^{0.5})/2\] But we need a unique solution, therefore, depending upon the condition x>=-7/2 you have to consider whether to take solution with -ve sign or +ve sign.

OpenStudy (anonymous):

Did you get it or do I need to give another hint for final step?

OpenStudy (anonymous):

yeah i don't get that part:/

OpenStudy (anonymous):

Considering that you understood how I got hold of x in term of y, you can observe that\[x=-7/2\pm (\lambda); \lambda>=0\] But as x>=-7/2, our solution would be correct when we take positive sign i.e.\[x=-7/2+\lambda\] because for unique inverse, we had got to eliminate one of the sign or our solution won't have been unique but inverse is a unique function. QED

OpenStudy (anonymous):

wait what does the \[\lambda \] mean?

OpenStudy (anonymous):

\[\lambda=(49+2y)^{0.5}/2\]

OpenStudy (anonymous):

so your saying the answer would be -7/2 + (49+2y)^(0.5)/2?

OpenStudy (anonymous):

oh wait do i still need to distribute the .5 power and divide?

OpenStudy (anonymous):

no this the final solution unless you want to make use of binomial theorem to distribute 0.5 power as there is no canonical expression for distribution of power of 0.5

OpenStudy (anonymous):

but aren't we supposed to solve for y so why would we leave a y in the answer?

OpenStudy (anonymous):

Just interchange the variable x and y to get y in terms of x and that would be your answer.

OpenStudy (anonymous):

got it! thank you!

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