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3 years ago

Use implicit differentiation to find the negative slope of a tangent to the circle x^2+y^2=16 at x=-2

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

3 0

Answer:

slope = - $\frac{\sqrt{3} }{3}$

Step-by-step explanation:

Differentiating implicitly with respect to x

2x + 2y $\frac{dy}{dx}$ = 0

2y $\frac{dy}{dx}$ = - 2x

$\frac{dy}{dx}$ = - $\frac{2x}{2y}$ = - $\frac{x}{y}$

$\frac{dy}{dx}$ is the measure of the slope of the tangent

rearrange equation to find corresponding y-coordinate of x = - 2

y² = 16 - 4 = 12 = 2 $\sqrt{3}$ ⇒ y = ± 2 $\sqrt{3}$

using x = - 2, y = - 2 $\sqrt{3}$ , then

$\frac{dy}{dx}$ = - $\frac{1}{\sqrt{3} }$ = - $\frac{\sqrt{3} }{3}$

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Antonio’s pizza charge $8.00 for a large pizza and $1.50 for each topping zine’s pizzeria charge $10.00 for a large pizza and $1

hichkok12 [17]

In this problem you would set these two equations equal to each other.
$8+ $1.50x = $10+ $1x
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3 years ago

A line that includes the point (7, -2) has a slope of 1. What is its equation in slope-intercept

kompoz [17]

Answer:

Y = x - 9

Step-by-step explanation:

A slope of 1 = x

So we have

Y = x

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Y = x - 9

6 0

3 years ago

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre

Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = $\frac{d}{dx}$ [ $x^{4}ln(x)$ ]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = $4x^{3}ln(x) + x_{4}.\frac{1}{x}$

f'(x) = $4x^{3}ln(x) + x^{3}$

f'(x) = $x^{3}[4ln(x) + 1]$

Now, find the critical points: f'(x) = 0

$x^{3}[4ln(x) + 1]$ = 0

$x^{3} = 0$

x = 0

and

$4ln(x) + 1 = 0$

$ln(x) = \frac{-1}{4}$

x = $e^{\frac{-1}{4} }$

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval x-value f'(x) result

0<x<0.78 0.5 f'(0.5) = -0.22 decreasing

x>0.78 1 f'(1) = 1 increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = $x^{4}ln(x)$

f(0.78) = $0.78^{4}ln(0.78)$