Find the greatest maximum y-value of the following equation:

1 answer:

5 0

A quick way to do this is by recalling the fact that $-1\le\cos x\le1$ , so the largest value that $\cos ax$ alone (for any real value of $a$ ) can take on is 1.

This means $-3\le3\cos2x\le3$ , and so $1\le3\cos2x+4\le7$ .

So the maximum value of this function is 7.

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Factor both quadratic expressions. (x 4 + 5x 2 - 36)(2x 2 + 9x - 5) = 0

Sonbull [250]

So focusing on x^4 + 5x^2 - 36, we will be completing the square. Firstly, what two terms have a product of -36x^4 and a sum of 5x^2? That would be 9x^2 and -4x^2. Replace 5x^2 with 9x^2 - 4x^2: $x^4+9x^2-4x^2-36$

Next, factor x^4 + 9x^2 and -4x^2 - 36 separately. Make sure that they have the same quantity inside of the parentheses: $x^2(x^2+9)-4(x^2+9)$

Now you can rewrite this as $(x^2-4)(x^2+9)$ , however this is not completely factored. With (x^2 - 4), we are using the difference of squares, which is $a^2-b^2=(a+b)(a-b)$ . Applying that here, we have $(x+2)(x-2)(x^2+9)$ . x^4 + 5x^2 - 36 is completely factored.

Next, focusing now on 2x^2 + 9x - 5, we will also be completing the square. What two terms have a product of -10x^2 and a sum of 9x? That would be 10x and -x. Replace 9x with 10x - x: $2x^2+10x-x-5$