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professor190 [17]

2 years ago

5

Help plzzzSelect the correct answer.

1 answer:

Morgarella [4.7K]2 years ago

3 0

<h3>Answer: C) 3</h3>

=========================================================

Explanation:

f(x) is the outer function, so the final output -8 corresponds to f(x)

We see that f(-4) = -8 in the first column of the table. I'm starting with the output and working my way backward to get the input. So we started with -8 and worked back to -4.

Then we move to the g(x) function to follow the same pattern: start with the output and move to the input. We start at -4 in the g(x) bubble and move to 3 in the x bubble.

In short, g(3) = -4

So,

f(g(x)) = f(g(3)) = f(-4) = -8

We see that x = 3 leads to f(g(x)) = -8

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For f(x)=4sin(x2) between x=0 and x=3, find the coordinates of all intercepts, critical points, and inflection points to two dec

telo118 [61]

Answer:

Intercepts:

x = 0, y = 0

x = 1.77, y = 0

x = 2.51, y = 0

Critical points:

x = 1.25, y = 4

x = 2.17 , y = -4

x = 2.8, y = 4

Inflection points:

x = 0.81, y = 2.44

x = 1.81, y = -0.54

x = 2.52, y = 0.27

Step-by-step explanation:

We can find the intercept by setting f(x) = 0

$4sin(x^2) = 0$

$sin(x^2) = 0$

$x^2 = n\pi$ where n = 0, 1, 2,3, 4, 5,...

$x = \sqrt(n\pi)$

Since we are restricting x between 0 and 3 we can stop at n = 2

So the function f(x) intercepts at y = 0 and x:

x = 0

x = 1.77

x = 2.51

The critical points occur at the first derivative = 0

$f^{'}(x) = 4cos(x^2)2x = 8xcos(x^2) = 0$

$xcos(x^2) = 0$

$x = 0$ or

$cos(x^2) = 0$

$x^2 = \frac{\pi}{2} + n\pi$ where n = 0, 1, 2, 3

$x = \sqrt{\pi(n+1/2)}$

Since we are restricting x between 0 and 3 we can stop at n = 2

So our critical points are at

x = 1.25, $y = f(1.25) = 4sin(1.25^2) = 4$

x = 2.17 , $y = f(2.17) = 4sin(2.17^2) = -4$

x = 2.8, $y = f(2.8) = 4sin(2.8^2) = 4$

For the inflection point, we can take the 2nd derivative and set it to 0

$f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0$

$cos(x^2) = 2x^2sin(x^2)$

$tan(x^2) = \frac{1}{2x^2}$

We can solve this numerically to get the inflection points are at

x = 0.81, $y = f(0.81) = 4sin(0.81^2) = 2.44$

x = 1.81, $y = f(1.81) = 4sin(1.81^2) = -0.54$

x = 2.52, $y = f(2.52) = 4sin(2.52^2) = 0.27$

3 0

2 years ago