Maths Inequality questions

anpanman SG

1. Sketch, on the same diagram, the graphs of y = (|x-5| - |2-3x|) and y = 7-4x. Hence or otherwise, solve the inequality ln (|x-5| - |2-3x|) </= ln (7-4x)

I was able to sketch the graphs. However the answer for the 2nd part is -1.5 <x< 1.75. From the graph, I know that beyond -1.5, (|x-5| - |2-3x|)| is STILL less than 7-4x graph. And beyond 1.75, (until x=2.83), did the graph (|x-5| - |2-3x|) become greater than 7-4x. May I know why is this so? My teacher said Ln graph CANNOT take into consideration y<0 but that should be for the e^x graph isnt it? If you draw ln x graph, there are still -ve y values.

Anyway, let's assume what my teacher said is right, shoudnlt the answer be -1.5 </= x </= 1.75. Why exclude -1.5 and 1.75?

2. the function f is defined by

f:x |-> ln (8-x)^3, x<8

Solve the inequality ln (8-x)^3 </= x

Please guide me in this question. THanks

wee_ws SG

Hi,

For Q1, I'm not sure what function it is when you typed

y = |(x-5) - |2-3x)|.

Do you mean

y = |x - 5| - |2 - 3x|?

Also, I do not understand LHS of your inequality, i.e.

ln |(x-5) - |2-3x)|.

Do you mean ln [ |x - 5| - |2 - 3x| ]?

For Q2, let y = f(x) = 3 ln (8 - x), where x < 8. Consider the intersection between y = f(x) and y = x, where the x-coordinate may be found with GC.

Thanks.

Cheers,
Wen Shih

anpanman SG

Hi,

For Q1, I'm not sure what function it is when you typed

y = |(x-5) - |2-3x)|.

Do you mean

y = |x - 5| - |2 - 3x|?

Also, I do not understand LHS of your inequality, i.e.

ln |(x-5) - |2-3x)|.

Do you mean ln [ |x - 5| - |2 - 3x| ]?

For Q2, let y = f(x) = 3 ln (8 - x), where x < 8. Consider the intersection between y = f(x) and y = x, where the x-coordinate may be found with GC.

Thanks.

Cheers,
Wen Shih

Oops serious typo. I edited the entire question. Thanks for the help!

wee_ws SG

Hi,

For the first part of Q1, we note that y = |x - 5| - |2 - 3x| coincide with y = 7 - 4x for all values of x >= 2/3. The solution to the inequality is therefore x >= -3/2.

For the second part, we consider only those parts where both graphs are strictly above the x-axis, i.e. -3/2 < x < 7/4, since the natural log is defined only for positive values.

In particular, |x - 5| - |2 - 3x| = 0 when x = -3/2 and x = 7/4, but then ln [ |x - 5| - |2 - 3x| ] will not be defined. Hence, we must exclude these two x-values. Thanks for sharing a nice question.

Cheers,
Wen Shih