Let h(x) = |kx+5|, domain of f(x) is [-5,7] , domain of f(h(x)) is [-6,1] and range of h(x) is the same as the domain f(x), then the value of k is

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Let h(x) = |kx+5|, domain of f(x) is [-5,7] , domain of f(h(x)) is [-6,1] and range of h(x) is the same as the domain f(x), then the value of k is

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is the answer \(k\in [-\frac{1}{3},~~2]\) ?
its an mcq and options are 1/3 4/5 1 none

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and its a single correct
`domain of f(h(x)) is [-6,1] ` |dw:1440520009960:dw|
`domain of f(x) is [-5,7]` therefore we must have : `-5 < h(x) < 7` does that look fine so far ?
yep
h(x) is absolute value function, so we don't need to wry about the lower bound -5
thus 0
we just need to make sure that h(x) is not greater than 7 in the interval [-6, 1]
because f(x) becomes undefined for values greater than 7 |dw:1440520422396:dw|
what if we try to draw the graph of h(x) with the vertex -5/k and y intercept (0,5) and draw two legs of the mod function and try to forcefully get the coordinates (1,7) from one part and from other ( -6,7) and take their intersection?
I am still getting a range.. wbu?
same http://www.wolframalpha.com/input/?i=solve+%7C-6k%2B5%7C+%3C+7%2C+%7C-k%2B5%7C%3C7

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