Question 20 Only, Please show your work. Solve the problem.

Mathematics

1 answer:

Sveta_85 [38]3 years ago

8 0

Answer:

7 inches

Step-by-step explanation:

The area of a square is given by the formula ...

A = s² . . . . . s = side length

Using the given area, you have ...

49 in² = s² . . . . . fill in given value of area

7 in = s . . . . . . . take the positive square root of both sides

The length of a side of the square is 7 inches.

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Given f'(x) = (x-4)(4-2x), find the x-coordinate for relative maximum on the graph of f(x)

antiseptic1488 [7]

Answer:

The coordinate of the relative maximum is x=4.

Step-by-step explanation:

Given that the derivative of the function $f(x)$ is $f'(x) = (x-4)(4-2x)$ , the maxima and minima or the critical points can be found where $f'(x)=0,$ that is:

$f'(x) = (x-4)(4-2x)=0.$

The solutions to this equation are $x=4$ and $x=2.$

Now, if the second derivative $f''(x)$ for a function $f(x)$ is negative at a critical point, then the critical point is the relative maximum.

Therefore we want to see at which of the two critical points is $f''(x)$ negative. The second derivative is:

$\frac{\mathrm{d f'(x)}}{\mathrm{d}x}=f''(x)=-4(x-3).$

Now $f''(4)$ is $-4$ , and $f''(2)$ is $+4$ , therefore we deduce that the relative maxium is located at $x=4$ , because there the second derivative $f''(x)$ is negative.