Do you even comprehend what calculus is? that's an algebra undertaking, undeniable and easy. Im happy i'd desire to be of help. Oh? Did you propose which you're in calculus... subsequently any question you are able to desire to probably have would desire to be a calculus question? Nope. Sorry to disappoint and insult your ego. common actuality, you arent waiting for calculus and in all probability shouldnt have handed algebra. comprehend what your question is earlier you bypass falsely labeling it and drawing the incorrect crowd.
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h^-1 (x) is y if h(y)=x
So we need to write x=h(y)=ln(4-19y).
Knowing x we want to solve for y (i.e., find y)
ln(4-19y)=x means 4-19y=e^x, where e is the usual constant 2.718282 (approx)
Subtract 4 from both sides
-19y=e^x-4
Divide by -19 on both sides
y= (e^x-4)/(-19)=(4-e^x)/19
Thus h^-1(x)=(4-e^x)/19
Hope this helps.
step one is to solve for x in terms of y. This involves manipulating ln(x)=e^x
y=ln(4-19x)=ln(4)/ln(-19x)=e^4/e^-19x
so e^-19x/e^4/y=1/y which implies that y=e^19x-4, now take the natural log of both sides
ln(y) = 19x-4 so x=(ln(y)+4)/19
step 2 is to change the y to x,so the inverse function is y=(ln(x)+4)/19
Let h(x) = y, then h^(-1)(y) = x
So, ln(4 - 19x) = y
4 - 19x = e^y
19x = 4 - e^y
x = (1/19)(4 - e^y)
Thus, h^(-1)(y) = (1/19)(4 - e^y), so h^(-1)(x) = (1/19)(4 - e^x)
Do you even comprehend what calculus is? that's an algebra undertaking, undeniable and easy. Im happy i'd desire to be of help. Oh? Did you propose which you're in calculus... subsequently any question you are able to desire to probably have would desire to be a calculus question? Nope. Sorry to disappoint and insult your ego. common actuality, you arent waiting for calculus and in all probability shouldnt have handed algebra. comprehend what your question is earlier you bypass falsely labeling it and drawing the incorrect crowd.