Math  /  Algebra

QuestionSolve the inequality. (Enter your answer using interval notation.) x2ex7ex<0x^{2} e^{x}-7 e^{x}<0

Studdy Solution

STEP 1

What is this asking? We need to find all xx values that make x2ex7exx^2 e^x - 7e^x less than zero. Watch out! Don't forget that exe^x is *always* positive!

STEP 2

1. Simplify the inequality
2. Analyze the inequality
3. Determine the solution interval

STEP 3

We've got x2ex7ex<0x^2 e^x - 7e^x < 0.
Since exe^x is in both terms, let's **factor it out**: ex(x27)<0e^x(x^2 - 7) < 0.
This makes things look much tidier!

STEP 4

Now, remember that exe^x is *always* **positive**, no matter what xx is.
So, we can **divide both sides** of our inequality by exe^x without flipping the inequality sign.
This gives us x27<0x^2 - 7 < 0.
Much simpler, right?

STEP 5

We have x27<0x^2 - 7 < 0.
Let's **rewrite** this as x2<7x^2 < 7.
This tells us that x2x^2 needs to be *smaller* than **7**.

STEP 6

To solve for xx, we'll **take the square root** of both sides.
Remember, when we do this, we get *two* possible solutions: a positive and a negative one.
So, we get 7<x<7-\sqrt{7} < x < \sqrt{7}.
This means xx has to be between 7-\sqrt{7} and 7\sqrt{7}.

STEP 7

Our solution is all the xx values between 7-\sqrt{7} and 7\sqrt{7}.
In **interval notation**, we write this as (7,7)(-\sqrt{7}, \sqrt{7}).
And that's it!

STEP 8

The solution to the inequality x2ex7ex<0x^2 e^x - 7e^x < 0 is (7,7)(-\sqrt{7}, \sqrt{7}).

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