12 equals x times the square root of 2

Mathematics

1 answer:

garik1379 [7]3 years ago

4 0

Answer:

x = 6√2

Step-by-step explanation:

Divide by the coefficient of x, then rationalize the denominator.

$12=x\sqrt{2} \qquad\text{given}\\\\\dfrac{12}{\sqrt{2}}=x \qquad\text{divide by $\sqrt{2}$}\\\\\dfrac{12}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=x \qquad\text{rationalize the denominator}\\\\\boxed{x=6\sqrt{2}} \qquad\text{simplify}$

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Graph y = 3(x + 2)3 - 3 and describe the end behavior.

-BARSIC- [3]

1) The function is 3(x + 2)³ - 3

2) The end behaviour is the limits when x approaches +/- infinity.

3) Since the polynomial is of odd degree you can predict that the ends head off in opposite direction. The limits confirm that.

4) The limit when x approaches negative infinity is negative infinity, then the left end of the function heads off downward (toward - ∞).

5) The limit when x approaches positive infinity is positivie infinity, then the right end of the function heads off upward (toward + ∞).

6) To graph the function it is important to determine:
- x-intercepts
- y-intercepts
- critical points: local maxima, local minima, and inflection points.

7) x-intercepts ⇒ y = 0

⇒ 3(x + 2)³ - 3 = 0 ⇒ (x + 2)³ - 1 = 0


⇒ (x + 2)³ = -1 ⇒ x + 2 = 1 ⇒ x = - 1


8) y-intercepts ⇒ x = 0

y = 3(x + 2)³ - 3 = 3(0 + 2)³ - 3 = 0 - 3×8 - 3 = 24 - 3 = 21


9) Critical points ⇒ first derivative = 0


i) dy / dx = 9(x + 2)² = 0


⇒ x + 2 = 0 ⇒ x = - 2


ii) second derivative: to determine where x = - 2 is a local maximum, a local minimum, or an inflection point.


y'' = 18 (x + 2); x = - 2 ⇒ y'' = 0 ⇒ inflection point.


Then the function does not have local minimum nor maximum, but an inflection point at x = -2.


Using all that information you can graph the function, and I attache the figure with the graph.