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3 years ago

At which root does the graph of f(x) = (x + 4)^6(x + 7)^5 cross the x axis?

2 answers:

7 0

ANSWER

The graph of
$f(x) = {(x + 4)}^{6} {(x + 7)}^{5}$
crosses the x-axis at
$x = - 7$

EXPLANATION

The nature of the multiplicity of a given polynomial function determines whether the graph crosses the x-axis at that intercept or not?

$f(x) = {(x + 4)}^{6} {(x + 7)}^{5}$

If the multiplicity of the factor is even as in
${(x + 4)}^{6}$
the graph touches but does not cross the x-axis at the intercept where
$x = - 4$
This means that the x-axis is a tangent to the function at this point.

However, if the multiplicity is odd, as in
$({x + 7})^{5}$
the graph crosses the x-axis at the intercept where
$x = - 7$

stepladder [879]3 years ago

6 0

Answer:

The Graph crosses at x -7

Step-by-step explanation:

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2 years ago

A company manufactures running shoes and basketball shoes. The total revenue (in thousands of dollars) from x1 units of running

Alborosie

Answer:

$x_1 =2 , x_2=7$

Step-by-step explanation:

Consider the revenue function given by $R(x_1,x_2) = -5x_1^2-8x_2^2 -2x_1x_2+34x_1+116x_2$ . We want to find the values of each of the variables such that the gradient( i.e the first partial derivatives of the function) is 0. Then, we have the following (the explicit calculations of both derivatives are omitted).

$\frac{dR}{dx_1} = -10x_1-2x_2+34 =0$

$\frac{dR}{dx_2} = -16x_2-2x_1+116 =0$

From the first equation, we get, $x_2 = \frac{-10x_1+34}{2}$ .If we replace that in the second equation, we get

$-16\frac{-10x_1+34}{2} -2x_1+116=0= 80x_1-2x_1+116-272= 78x_1-156$

From where we get that $x_1 = \frac{156}{78}=2$ . If we replace that in the first equation, we get

$x_2 = \frac{-10\cdot 2 +34}{2}=\frac{14}{2} = 7$

So, the critical point is $(x_1,x_2) = (2,7)$ . We must check that it is a maximum. To do so, we will use the Hessian criteria. To do so, we must calculate the second derivatives and the crossed derivatives and check if the criteria is fulfilled in order for it to be a maximum. We get that

$\frac{d^2R}{dx_1dx_2}= -2 = \frac{d^2R}{dx_2dx_1}$

$\frac{d^2R}{dx_{1}^2}=-10, \frac{d^2R}{dx_{2}^2}=-16$

We have the following matrix,

$\left[\begin{matrix} -10 & -2 \\ -2 & -16\end{matrix}\right]$ .

Recall that the Hessian criteria says that, for the point to be a maximum, the determinant of the whole matrix should be positive and the element of the matrix that is in the upper left corner should be negative. Note that the determinant of the matrix is $(-10)\cdot (-16) - (-2)(-2) = 156>0$ and that -10<0. Hence, the criteria is fulfilled and the critical point is a maximum

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