How do I solve for the blank

Mathematics

1 answer:

OLga [1]3 years ago

3 0

Look at the problem and do it in order

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Use the definition of continuity and the properties of limit to show that the function f(x)=x sqrtx/ (x-6)^2 is continuous at x=

jasenka [17]

Answer:

The function $\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$ is continuous at x = 36.

Step-by-step explanation:

We need to follow the following steps:

The function is:

$\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$

The function is continuous at point x=36 if:

The function $\\ f(x)$ exists at x=36.
The limit on both sides of 36 exists.
The value of the function at x=36 is the same as the value of the limit of the function at x = 36.

Therefore:

The value of the function at x = 36 is:

$\\ f(36) = \frac{36*\sqrt{36}}{(36-6)^{2}}$

$\\ f(36) = \frac{36*6}{900} = \frac{6}{25}$

The limit of the $\\ f(x)$ is the same at both sides of x=36, that is, the evaluation of the limit for values coming below x = 36, or 33, 34, 35.5, 35.9, 35.99999 is the same that the limit for values coming above x = 36, or 38, 37, 36.5, 36.1, 36.01, 36.001, 36.0001, etc.

For this case:

$\\ lim_{x \to 36} f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$

$\\ \lim_{x \to 36} f(x) = \frac{6}{25}$

Since

$\\ f(36) = \frac{6}{25}$

And

$\\ \lim_{x \to 36} f(x) = \frac{6}{25}$

Then, the function $\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$ is continuous at x = 36.

8 0

3 years ago

Find the vertex, focus, and directrix. y = 1/24(x+1)² - 3.

Vedmedyk [2.9K]

$y = \frac{1}{24}(x+1)^2 - 3\\\\y+3 =\frac{1}{24}(x+1)^2\ \ / *24\\\\ (x+1)^2 = 24(y+3)$

This is an equation of a parabola that opens upwards.

$Its \ standard \ form: \\(x-h)^2=4p(y-k)\\ (h,k)=(x,y) \ coordinates \ of \ the \ vertex\\\ (h,k)=(-1,-3) \\\\axis \ of \ symmetry: \ x= -1\\ \\4p=24\ \ /:4\\p=6$

$focus:(h,k+p)=(-1,-3+6)=(-1,3) \\ \\directrix: \ y=k-p=-3-6=-9$