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andrew-mc [135]
4 years ago
10

|4x-3|-5=4 and |2x+3|+8=3, what are the solutions to the problems

Mathematics
1 answer:
horsena [70]4 years ago
3 0

|4x-3|-5=4 \\|4x-3|=9\\4x-3=9 \vee 4x-3=-9\\4x=12 \vee 4x=-6\\x=3 \vee x=-\dfrac{3}{2}

|2x+3|+8=3\\|2x+3|=-5\\x\in\emptyset

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Illusion103
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Use the graph of f '(x) below to find the x values of the relative maximum on the graph of f(x):
Lana71 [14]

Answer:

You have relative maximum at x=1.

Step-by-step explanation:

-Note that f' is continuous and smooth everywhere. f therefore exists everywhere on the domain provided in the graph.

f' is greater than 0 when the curve is above the x-axis.

f' greater than 0 means that f is increasing there.

f' is less than 0 when the curve is below the x-axis.

f' is less than 0 means that f is decreasing there.

Since we are looking for relative maximum(s), we are looking for when the graph of f switches from increasing to decreasing. That forms something that looks like this '∩' sort of.

This means we are looking for when f' switches from positive to negative. At that switch point is where we have the relative maximum occurring at.

Looking at the graph the switch points are at x=0, x=1, and x=2.

At x=0, we have f' is less than 0 before x=0 and that f' is greater than 0 after x=0.  That means f is decreasing to increasing here. There would be a relative minimum at x=0.

At x=1, we have f' is greater than 0 before x=1 and that f' is less than 0 after x=1. That means f is increasing to decreasing here. There would be a relative maximum at x=1.

At x=2, we have f' is less than 0 before x=2 and that f' is greater than 0 after x=2. That means f is decreasing to increasing here. There would be a relative minimum at x=2.

Conclusion:

* Relative minimums at x=0 and x=2

* Relative maximums at x=1

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