Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in interval notation.) Correct: Your answer is correct. Find the interval on which f is decreasing. (Enter your answer in interval notation.) Correct: Your answer is correct. (b) Find the local minimum and maximum values of f. local minimum -1519 Correct: Your answer is correct. local maximum 1856 Correct: Your answer is correct. (c) Find the inflection point.

(a). The function is increasing in the interval:

The function is decreasing in the interval:

(b). The local maximum value is

The local minimum value is

(c). The inflection point is .

We have the given function:

Differentiating it with respect to x, we get

Put first derivative equal to zero

(a). Now, the function is increasing in the interval:

and the function is decreasing in the interval:

(b). Now, since   the function has local maxima at x=-8 and the local maximum value is

Also since the function has local minima at x=7 and the local minimum value is

(c). For the inflection point put second derivative equal to zero .

So, the inflection point is .

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We have been given a function . We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

So are critical points and these will divide our function in 3 intervals .

Now we will find derivative over each interval as:

Since , therefore, function is increasing on interval .

Since , therefore, function is decreasing on interval .

Let us check for the derivative at .

Since , therefore, function is increasing on interval .

(b) Since are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

Therefore, is a local maximum and is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

Therefore, is an inflection point of given function.

but there’s no any figure there!

the value of x is 9 ⇒ answer b

step-by-step explanation:

* we can solve this problem using cosine and sine rule

- in δ abc

∵ ab = 36 , bc = 28 , ac = 16

- lets find the measure of angle b using the cosine rule

∵ cos b = [ab² + bc² - ac²]/2(ab)(bc)

∴ cos b = [36² + 28² - 16²]/2(36)(28)

∴ cos b = 19/21

∴ m∠b = cos^-1(19/21) ≅ 25°

- lets find the measure of angle a using the sine rule

∵ sina/bc = sinb/ac = sinc/ab

∴ sina/28 = sin25°/16 ⇒ by using cross multiplication

∴ sina = 28 × sin25°/16

∴ sina = 0.7396

∴ m∠a = sin^-1(0.7396) ≅ 48°

- from the figure m∠abd = m∠dbc

∵ m∠abc = 25°

∴ m∠abd = 25°/2 = 12.5°

- in δabd

∵ m∠a = 48° , m∠abd = 12.5°

∵ the sum of the measures of the interior angles of any δ is 180°

∴ m∠adb = 180° - (48° + 12.5°) = 119.5°

∵ ab = 36

∵ ad = x

- use the sine rule to find x

∵ sin∠adb/ab = sin∠abd/x

∴ sin119.5°/36 = sin12.5°/x ⇒ by using cross multiplication

∴ x = 36 × sin12.5°/sin119.5° = 8.95 ≅ 9

* the value of x is 9