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8. Given the following​ function, (a) find the​ vertex; (b) determine whether there is a maximum or a minimum​ value, and find the​ value; (c) find the​ range; and​ (d) find the intervals on which the function is increasing and the intervals on which the function is decreasing.

8 Given the following function a find the vertex b determine whether there is a maximum or a minimum value and find the value c find the range and d find the in class=

Respuesta :

devishri1977

Answer:

Step-by-step explanation:

Parabola:

   [tex]\sf f(x) =\dfrac{-1}{2}x^2+7x - 5\\\\a = \dfrac{-1}{2} \ ; b = 7 \ ; \ c = -5[/tex]

[tex]\sf x = \dfrac{-b}{2a}\\\\x =\dfrac{-7}{2*\dfrac{-1}{2}}\\\\ = \dfrac{-7}{-1}[/tex]

x = 7

Now, to find the y-value substitute this in the equation

               [tex]\sf f(x) = \dfrac{-1}{2}*7^2+7*7-5\\\\ =\dfrac{-49}{2}+49-5\\\\ = \dfrac{-49}{2}+44\\\\ = \dfrac{-49}{2}+\dfrac{88}{2}\\\\= \dfrac{39}{2}[/tex]

[tex]\boxed{Vertex(7 ,\dfrac{39}{2})}[/tex]

b) The value of a is negative. So, the parabola is open down words and the maximum value is given by the y-coordinate of the Vertex.

             [tex]\sf \boxed{Maximum \ value= \dfrac{39}{2}}[/tex]

[tex]\sf c) Range = [\dfrac{39}{2}, -infinity)[/tex]

VectorFundament120

Answer:

(a) (7,39/2)

(b) Maximum, 39/2

(c) y ≤ 39/2

(d) Increase when x < 7 and decrease when x > 7

Step-by-step explanation:

Given the parabola:

[tex]\displaystyle \large{f(x)=-\dfrac{1}{2}x^2+7x-5}[/tex]

( A ) Find the vertex

In order to find the vertex, let’s use calculus for this one. Recall the power rules for differentiation:

Power Rules

[tex]\displaystyle \large{f(x)=x^n \to f'(x)=nx^{n-1}}\\\\\displaystyle \large{f(x)=kx^n \to f'(x)=knx^{n-1} \quad \tt{(k \ \ is \ \ constant)}}\\\\\displaystyle \large{f(x)=k \to f'(x)=0 \quad \tt{(k \ \ is \ \ constant)}}[/tex]

Derivative Definition

  • Derivative f'(x) is a slope itself or rate of changes.

Derive the parabola:

[tex]\displaystyle \large{f'(x)=-\dfrac{1}{2}(2)x^{2-1} + 7(1)x^{1-1} -0}\\\\\displaystyle \large{f'(x)=-x+7}[/tex]

Since vertex has slope = 0 —> e.g f'(x) = 0:

[tex]\displaystyle \large{0=-x+7}\\\\\displaystyle \large{-x=-7}\\\\\displaystyle \large{x=7}[/tex]

Substitute x = 7 in f(x):

[tex]\displaystyle \large{f(7)=-\dfrac{1}{2}(7)^2+7(7)-5}\\\\\displaystyle \large{f(7)=-\dfrac{1}{2}(49)+49-5}\\\\\displaystyle \large{f(7)=-\dfrac{49}{2}+44}\\\\\displaystyle \large{f(7)=-\dfrac{49}{2}+\dfrac{88}{2}}\\\\\displaystyle \large{f(7)=\dfrac{39}{2}}[/tex]

Therefore, the vertex is at (7,39/2)

( B ) Determine if max or min then find the value

Since the parabola opens downward then there only exists maximum value. The maximum value is the y-value of vertex at x-value of vertex. Henceforth:

  • There is maximum value but no minimum value and the maximum value is 39/2 at x = 7.

( C ) Find range

For parabola, range is minimum value </≤ y </≤ maximum value. We know that parabola has maximum value of 39/2 but no minimum value so we can just ignore it then we’d have:

  • y ≤ 39/2 is our range

[Note: < and > are for open-dot meaning the value will not be included —> e.g x > 4 means 4 isn’t included in.]

( D ) Find the interval when function is increasing and when it’s decreasing

For parabola, the function will increase only if f'(x) or slope > 0 and will decrease only f'(x) < 0.

We know, from part A that the derivative is:

[tex]\displaystyle \large{f'(x)=-x+7}[/tex]

Therefore, when f'(x) > 0 —> e.g -x + 7 > 0:

[tex]\displaystyle \large{-x+7 > 0}\\\\\displaystyle \large{-x > -7}\\\\\displaystyle \large{x < 7}[/tex]

When f'(x) < 0 —> e.g -x + 7 < 0:

[tex]\displaystyle \large{-x+7 < 0}\\\\\displaystyle \large{-x < -7}\\\\\displaystyle \large{x > 7}[/tex]

Therefore, the function will increase when x < 7 and will decrease when x > 7.

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