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wolverine [178]
3 years ago
9

K=1; f(x)=4x^3-2x^2+2x+4; Lower bound?

Mathematics
1 answer:
photoshop1234 [79]3 years ago
4 0
Given:
f(x) = 4x^{3} - 2x^{2} + 2x + 4
and,
K = 1

We would require to do the synthetic division of the above problem in order to know whether K=1 is a lower bound or not. Let's do it!

Steps of Synthetic Division:

1. Write the coefficients of the equation, and before that write 1. Like,

1 ║ 4   -2   2   4

As you can see, the coefficients of the equation are seperated by ║.

2. Drop down the first value after ║as it is. Like:
1 ║ 4   -2   2   4
        :
----------------------
       4

3. Multiply 4 with 1 and then  place the resultant value under -2 and then add that resultant value with -2. Like
1 ║ 4   -2   2   4
        :    4
----------------------
       4   2

4. Repeat Step 3 until you're done.

1 ║ 4   -2   2   4
        :    4   2   
----------------------
       4   2    4   

1 ║ 4   -2   2   4
        :    4   2   4
----------------------
       4   2    4   8


Now every value, +4, +2, +4, +8, under the bar(---------) is positive; therefore, it means that K=1 is the upper bound, NOT lower bound.

Ans: K=1 is NOT a lower bound.
-i
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Find the directional derivative of the function at the given point in the direction of the vector v. G(r, s) = tan−1(rs), (1, 3)
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The <em>directional</em> derivative of f at the given point in the direction indicated is \frac{5}{2}.

<h3>How to calculate the directional derivative of a multivariate function</h3>

The <em>directional</em> derivative is represented by the following formula:

\nabla_{\vec v} f = \nabla f (r_{o}, s_{o})\cdot \vec v   (1)

Where:

  • \nabla f (r_{o}, s_{o}) - Gradient evaluated at the point (r_{o}, s_{o}).
  • \vec v - Directional vector.

The gradient of f is calculated below:

\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{\partial f}{\partial r}(r_{o},s_{o})  \\\frac{\partial f}{\partial s}(r_{o},s_{o}) \end{array}\right]   (2)

Where \frac{\partial f}{\partial r} and \frac{\partial f}{\partial s} are the <em>partial</em> derivatives with respect to r and s, respectively.

If we know that (r_{o}, s_{o}) = (1, 3), then the gradient is:

\nabla f(r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{s}{1+r^{2}\cdot s^{2}} \\\frac{r}{1+r^{2}\cdot s^{2}}\end{array}\right]

\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{3}{1+1^{2}\cdot 3^{2}} \\\frac{1}{1+1^{2}\cdot 3^{2}} \end{array}\right]

\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{3}{10} \\\frac{1}{10} \end{array}\right]

If we know that \vec v = 5\,\hat{i} + 10\,\hat{j}, then the directional derivative is:

\nabla_{\vec v} f = \left[\begin{array}{cc}\frac{3}{10} \\\frac{1}{10} \end{array}\right] \cdot \left[\begin{array}{cc}5\\10\end{array}\right]

\nabla _{\vec v} f (r_{o}, s_{o}) = \frac{5}{2}

The <em>directional</em> derivative of f at the given point in the direction indicated is \frac{5}{2}. \blacksquare

To learn more on directional derivative, we kindly invite to check this verified question: brainly.com/question/9964491

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