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lisabon 2012 [21]
3 years ago
15

The function h(x) = x2 + 14x + 41 represents a parabola. Part A: Rewrite the function in vertex form by completing the square. S

how your work. (6 points) Part B: Determine the vertex and indicate whether it is a maximum or a minimum on the graph. How do you know? (2 points) Part C: Determine the axis of symmetry for h(x). (2 points)
Mathematics
1 answer:
denpristay [2]3 years ago
3 0
Given the function h(x)=x^2+14x+41, to solve by completing square we procced as follows;
x^2+14x+41=0
x^2+14x=-41
but;
c=(b/2)^2
and b=14
hence;
c=(14/2)^2=49
substituting the value of c in the expression we get:
x^2+14x+49=-41+49
x^2+14x+49=8
(x+7)^2=8
this can be written in vertex form;
h(x)=a(x-h)^2+k
where:
(h,k) is the vertex;
thus
(x+7)^2=8
h(x)=(x+7)^2-8
hence the vertex will be at the point:
(-7,-8)
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Find dy/dx for y= x^3 ln (cot x)
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<h3>Answer</h3>

  \dfrac{dy}{dx} = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)

<h3>Explanation</h3>

By the product rule (d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x), we have

  \begin{aligned}\frac{dy}{dx} &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}

By the chain rule:

  \begin{aligned}\big(\ln (\cot x)\big)' &= \dfrac{1}{\cot x} \cdot (\cot x)' \\ &= \dfrac{1}{\cot x} \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = - \frac{1}{\cos x} \cdot \frac{1}{\sin x} \\&= -\csc(x)\sec(x)\end{aligned}

By the power rule:

  (x^3)' = 3x^2

thus

  \begin{aligned}\frac{dy}{dx} &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}

Nothing to do to simplify any further, other than factoring out x^2.

4 0
3 years ago
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