# ASK YOUR TEACHERConsider the function below. f(x) = −2x2 + 12x + 2(a) Find the critical numbers of f. (Enter your answers as

###### Question:

## Answers

(a) The critical number of [tex]f(x)[/tex] are [tex]x=-4, 1[/tex]

(b)

Increasing for [tex](-\infty, -4)[/tex]Decreasing for [tex](-4, 1)[/tex]Increasing for [tex](1, \infty)[/tex](c)

relative maximum [tex](-4, 112)[/tex]relative minimum [tex](1, -13)[/tex]Step-by-step explanation:

(a) The critical numbers of a function are given by finding the roots of the first derivative of the function or the values where the first derivative does not exist. Since the function is a polynomial, its domain and the domain of its derivatives is [tex](-\infty, \infty)[/tex]. Thus:

[tex]\frac{df(x)}{dx} = \frac{d(2x^3+9x^2-24x)}{dx} =6 x^2+18x -24\\6 x^2+18x -24=0\\\boxed{x=-4, x=1}[/tex]

(b)

A function [tex]f(x)[/tex] defined on an interval is monotone increasing on [tex](a, b)[/tex] if for every [tex]x_1, x_2 \in (a, b): x_1[/tex] implies [tex]f(x_1)[/tex]A function [tex]f(x)[/tex] defined on an interval is monotone decreasing on [tex](a, b)[/tex] if for every [tex]x_1, x_2 \in (a, b): x_1[/tex] implies [tex]f(x_1)f(x_2)[/tex]Combining the domain [tex](-\infty, \infty)[/tex] with the critical numbers we have the intervals [tex](-\infty, -4)[/tex], [tex](-4, 1)[/tex] and [tex](1, \infty)[/tex]. Note that any of the points are included, in the case of the infinity it is by definition and the critical number are never included because the function monotony is not defined in the critical points, i.e. it is not monotone increasing or decreasing. Now, let's check for the monotony in each interval, for this, we check for the sign of the first derivative in each interval. Evaluating in each interval the first derivative (one point is enough), we obtain the monotony of the function to be:

Increasing for [tex](-\infty, -4)[/tex]Decreasing for [tex](-4, 1)[/tex]Increasing for [tex](1, \infty)[/tex](c) From the values obtained in (a) so the relative extremum are the points [tex](-4, 112)[/tex] and [tex](1, -13)[/tex]. The [tex]y[/tex]-values are found by evaluating the critical numbers in the original function. Since the first derivative decreases after passing through [tex]x=-4[/tex] and increases after passing through the point [tex]x=1[/tex] we have:

relative maximum [tex](-4, 112)[/tex]relative minimum [tex](1, -13)[/tex]a) DNE

b) The function increases for every real value of x.

c) DNE

Step-by-step explanation:

Given a function f(x), the critical points are those in which [tex]f^{\prime}(x) = 0[/tex], that is, the roots of the first derivative of f(x).

Those critical points let us find the intervals in which the function increases or decreases. If the first derivative in the interval is positive, the function increases in the interval. If it is negative, the function decreases.

If the function increases before a critical point and then, as it passes the critical point, it starts to decrease, we have that the critical point [tex](x_{c}, f(x_{c}) is a relative maximum.If the function decreases before a critical point and then, as it passes the critical point, it starts to increase, we have that the critical point [tex](x_{c}, f(x_{c}) is a relative minimum.If the function has no critical points, it either always increases or always decreases.In this exercise, we have that:[tex]f(x) = 7x + 1[/tex]

(a) Find the critical numbers of f.

[tex]f^{\prime}(x) = 0[/tex]

[tex]f^{/prime}(x) = 7[/tex]

7 = 0 is false. This means that f has no critical points.

(b) Find the open intervals on which the function is increasing or decreasing.

Since there are no critical points, we know that either the function increases or decreases in the entire real interval.

We have a first order function in the following format:

[tex]f(x) = ax + b[/tex]

In which [tex]a 0[/tex].

So the function increases for every real value of x.

(c) Apply the First Derivative Test to identify the relative extremum.

From a), we find that there are no critical numbers. So DNE

Increasing in x∈(0,π/4)∪(5π/4,2π) decreasing in(π/4,5π/4)

Step-by-step explanation:

given f(x) = sin(x) + cos(x)

f(x) can be rewritten as [tex]\sqrt{2} [\frac{sin(x)}{\sqrt{2} }+\frac{cos(x)}{\sqrt{2} } ]..................(a)\\\\\ \frac{1}{\sqrt{2} } = cos(45) = sin(45)\\\\[/tex]

Using these result in equation a we get

f(x) = [tex]\sqrt{2} [ cos(45)sin(x)+sin(45)cos(x)]\\\\= \sqrt{2} [sin(45+x)]..........(b)[/tex]

Now we know that for derivative with respect to dependent variable is positive for an increasing function

Differentiating b on both sides with respect to x we get

f '(x) = [tex]f '(x)=\sqrt{2} \frac{dsin(45+x)}{dx}\\ \\f'(x)=\sqrt{2} cos(45+x)\\\\f'(x)0=\sqrt{2} cos(45+x)0[/tex]

where x∈(0,2π)

we know that cox(x) > 0 for x∈[0,π/2]∪[3π/2,2π]

Thus for cos(π/4+x)>0 we should have

1) π/4 + x < π/2 => x<π/4 => x∈[0,π/4]

2) π/4 + x > 3π/2 => x > 5π/4 => x∈[5π/4,2π]

from conditions 1 and 2 we have x∈(0,π/4)∪(5π/4,2π)

Thus the function is decreasing in x∈(π/4,5π/4)

Increasing:

[tex](\frac{\pi}{3},\frac{5\pi}{3})[/tex]

Decreasing

[tex](0,\frac{\pi}{3})[/tex] U [tex](\frac{5\pi}{3},2\pi)[/tex]

Step-by-step explanation:

Increasing and Decreasing Intervals

To find if a function is increasing in a point x=a, we evaluate the first derivative in x=a and if:

f'(a) >0, the function is increasing f'(a) <0, the function is decreasing f'(a) =0, the function has a critical pointFor continuous functions, we can safely assume between a given critical point and the next one, the function keeps its behavior, i.e. it's increasing or decreasing in the interval formed by both points.

So, we find the critital points of

[tex]f(x)=x-2sinx[/tex]

Taking the derivative

[tex]f'(x)=1-2cosx[/tex]

Equating to 0

[tex]1-2cosx=0[/tex]

Solving

[tex]\displaystyle cosx=\frac{1}{2}[/tex]

There are two solutions in the interval [tex](0,2\pi)[/tex]

[tex]\displaystyle x=\frac{\pi}{3},\ x=\frac{5\pi}{3}[/tex]

Now we compute the second derivative

[tex]f''(x)=2sinx[/tex]

Evaluating for both critical points

[tex]\displaystyle f''(\frac{\pi}{3})=2sin\frac{\pi}{3}=\sqrt{3}[/tex]

Since it's positive, the point is a minimum

[tex]\displaystyle f''(\frac{5\pi}{3})=2sin\frac{5\pi}{3}=-\sqrt{3}[/tex]

Since it's negative, the point is a maximum

In the interval

[tex](0,\frac{\pi}{3})[/tex]

the function is decreasing

In the interval

[tex](\frac{\pi}{3},\frac{5\pi}{3})[/tex]

the function is increasing

In the interval

[tex](\frac{5\pi}{3},2\pi)[/tex]

the function is decreasing

Required critical point is (5, -25) which takes local minima.

Step-by-step explanation:

Given function is,

[tex]f(x)=x^2-10x\hfill (1)[/tex]

To find the critical point we have to do,[tex]f'(x)=0\implies 2x-10=0\implies x=5[/tex]

Now substitite this value in (1) we get,

[tex]y=f(x)=(5)^2-(10\times 5)=-25[/tex]

Therefore the critical point is (5,-25).

To examine critical value takes maxima or minima we have to consider double derivative of (1) at x=5, that is,[tex]f''(5)=20[/tex]

since double derivative at x=5 is geater than zero so critical point (5,-25) takes local minima.

There exist no any local maxima.I believe the answer is C