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2 years ago

The equation x^3-5x=59 has a solution between 4and 5 . Use trail and improvement method. Give your answer to 1 decimal place.

Mathematics

1 answer:

3241004551 [841]2 years ago

5 0

Answer:

4.3 to 1 decimal place.

Step-by-step explanation:

x^3 - 5x - 59 = 0

Try x = 4.1:

f(4.1) = (4.1)^3 - 5(4.1) - 59 = -10.579

Try: 4.2:

f(4.2) = 4.2^3 - 5(4.2) - 59 = -5.912, Try x = 4.3:

f(4.3) = -0.993 (getting closer to 0!!)

f(4.4) = 4.184

NOTICE - there is a change of sign in the result so the root lies between 4.3 and 4.4.

Try x = 4.35:

f(4.35) = 1.563 so it looks like the root is closer to 4.3 than 4.4 because 1.563 is closer to zero than 4.184.

Just to double check let's calculate x = 4.32:

f(4.32) = 0.0216 - which is very close to zero.

So the answer is 4.3 to 1 decimal place.

(I've checked it out on my calculator - the root is 4.31958).

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Which of the following rational function is graphed below?

steposvetlana [31]

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Answer:

B. 1/((x -1)(x -2))

Step-by-step explanation:

The vertical asymptotes at x=1 and x=2 tell you the denominator will have factors such that these values make it zero: (x -1)(x -2). That is sufficient to identify choice B as the correct answer.

$f(x)=\dfrac{1}{(x-1)(x-2)}$

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2 years ago

ANY JESUS HELPERS PLEASE HELP The incorrect work of a student to solve an equation 2(y + 4) = 4y is shown below: Step 1: 2(y + 4

wariber [46]

Answer:

2 should be distributed as 2y + 8; y = 4

Step-by-step explanation:

2(y + 4) = 4y

Distribute

2y + 8 = 4y

Step 2 is incorrect because the added instead of multiplied

Continuing on to correct the problem

Subtract 2y from each side

2y+8-2y = 4y-2y

8 = 2y

Divide by 2

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3 0

3 years ago

Read 2 more answers

What is 204 thousands

max2010maxim [7]

204 thousands equals to 204, 000 as the mathematical value and expression.

In expanded form
200, 000 + 4, 000 = 204, 000

In word form,
Two hundred and four thousand

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3 years ago

Which group shows 0.25 of the faces smiling?

Ivahew [28]

Answer:

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3 years ago

Find the critical points of the function f(x, y) = 8y2x − 8yx2 + 9xy. Determine whether they are local minima, local maxima, or

NARA [144]

Answer:

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

Step-by-step explanation:

The function is:

$f(x,y) = 8\cdot y^{2}\cdot x -8\cdot y\cdot x^{2} + 9\cdot x \cdot y$

The partial derivatives of the function are included below:

$\frac{\partial f}{\partial x} = 8\cdot y^{2}-16\cdot y\cdot x+9\cdot y$

$\frac{\partial f}{\partial x} = y \cdot (8\cdot y -16\cdot x + 9)$

$\frac{\partial f}{\partial y} = 16\cdot y \cdot x - 8 \cdot x^{2} + 9\cdot x$

$\frac{\partial f}{\partial y} = x \cdot (16\cdot y - 8\cdot x + 9)$

Local minima, local maxima and saddle points are determined by equalizing both partial derivatives to zero.

$y \cdot (8\cdot y -16\cdot x + 9) = 0$

$x \cdot (16\cdot y - 8\cdot x + 9) = 0$

It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

$\left \{ {{-16\cdot x + 8\cdot y=-9} \atop {-8\cdot x + 16\cdot y=-9}} \right.$

The solution of the system is (3/8, -3/8).

Let assume that y = 0, the nonlinear system is reduced to a sole expression:

$x\cdot (-8\cdot x + 9) = 0$

Another solution is (9/8,0).

Now, let consider that x = 0, the nonlinear system is now reduced to this:

$y\cdot (8\cdot y+9) = 0$

Another solution is (0, -9/8).

The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

$H = \frac{\partial^{2} f}{\partial x^{2}} \cdot \frac{\partial^{2} f}{\partial y^{2}} - \frac{\partial^{2} f}{\partial x \partial y}$

The second derivatives of the function are:

$\frac{\partial^{2} f}{\partial x^{2}} = 0$

$\frac{\partial^{2} f}{\partial y^{2}} = 0$

$\frac{\partial^{2} f}{\partial x \partial y} = 16\cdot y -16\cdot x + 9$

Then, the expression is simplified to this and each point is tested:

$H = -16\cdot y +16\cdot x -9$

S1: (0,0)

$H = -9$ (Saddle Point)

S2: (3/8,-3/8)

$H = 3$ (Local maximum or minimum)

S3: (9/8, 0)

$H = 9$ (Local maximum or minimum)

S4: (0, - 9/8)

$H = 9$ (Local maximum or minimum)

Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:

S2: (3/8,-3/8)

$f(\frac{3}{8} ,-\frac{3}{8} ) = - \frac{27}{64}$ (Local minimum)

S3: (9/8, 0)

$f(\frac{9}{8},0) = 0$ (Local maximum)

S4: (0, - 9/8)

$f(0,-\frac{9}{8} ) = 0$ (Local maximum)

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

4 0

3 years ago