Given f(x) = x-1/ 2, solve for f−1(5).
@Nnesha @mathstudent55
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replace f(x) with 5 and solve for x
so i got 11?
i mean 2
\[f^{-1}(5)=x \text{ \implies} f(x)=5\] \[5=\frac{x-1}{2} \\ \text{ multiply 2 on both sides } \\ 10=x-1\]
so -10?
so adding 1 on both sides does not give -10
11=x
10=x-1 add 1 to both sides yes 10+1=x 11=x this means since \[f(11)=5 \text{ then } f^{-1}(5)=11\]
I thought it would be 2 because 5-1/2 =2
why did you replace x with 5?
you were suppose to replace f(x) with 5 the question ask to find f inverse of 5 not f of 5 like this is the way I like to to think of it if it asked to find something like \[f^{-1}(5) \\ \text{ then I call this something like } x \\ f^{-1}(5)=x \\ \text{ but if } f^{-1}(5)=x \text{ then } f(x)=5 \\ \text{ so this means I can find the } x \text{ for when } f(x) \text{ is } 5 \\ \text{ by replace } f(x) \text{ with 5 and solving for } x \] \[f(x)=\frac{x-1}{2} \\ \text{ we are going \to replace } f(x) \text{ with 5, not } x \\ \text{ since we have }f(x)=5 \\5=\frac{x-1}{2} \\ \text{ now we can figure out the } x \text{ such that we do have } f(x)=5\] We figured out that that x was 11 so we have f(11)=5 so that means \[f^{-1}(5)=11 \]
Ohh! Okay :)
Which of the following is the conjugate of a complex number with 2 as the real part and −8 as the imaginary part?
i got 2+8i
correct
sounds fine
complex number: a + bi conjugate: a - bi As you can see , only the sign of the imaginary part changes to form the conjugate, which is what you did. Good job!
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