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(b)(i) range: (1 infinity)
(ii)domain of f^-1 is not same as domain of f; hence, we say that it has no inverse.
f^-1 = sqrt[(x-1)/2]
I am not sure about 4(a) (ii).
4(b) (i) taking ln, we get -1.73 lnx= ln4.82
lnx = ln4.82/-1.73 = -0.90912
x = e^-090912 =0.4029
(ii)taking ln, we get -xln2 = ln0.1
x = -ln0.1/ln2 = -1.4307
3.(b) (i) it can not represent as y =x^4+5 represents so that y is always above x-axis. the graph shown is below x-axis also. Hence, it can not represent given graph.
(ii) it represents the given graph.
(iii) it represents linear function but given graph is not linear; hence, it does not represent the given graph.
4.(a)(ii) x^-1 is answer.
it is not clear how numeric solution is possible.
I was not sure what R and Q are representing.
i will reply after 6-7 hours if there is any query.
5.(b)(i) range: (1 infinity)
5.(b)(i) Range is that y will vary from 1 to infinity. we can wriite range: 1≤y≤∞.
(ii) y = 2x^2+1
For getting inverse, we interchange, x and y and get
or y = sqrt[(x-1)/2]
Hence, f^-1 = sqrt[(x-1)/2]
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