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4 years ago

Find the point on the graph of the given function at which the slope of the tangent line is the given slope.

Mathematics

1 answer:

mezya [45]4 years ago

3 0

Answer:

A ( -2 , 9 )

Step-by-step explanation:

<u>Idea:</u> You find the first derivative of f(x), and then set it equal to the desired slope. You'll find some x. That we will use to find the point.

f'(x) = 3x^2 + 12x + 20

f'(x) = 8

3x^2 + 12x + 20 = 8

3x^2 + 12x + 12 = 0

3 ( x^2 + 4x + 4 ) = 0

3 ( x + 2 )^2 = 0

x + 2 = 0

x = -2

So, the desired point is:

A ( -2, f(-2) ) --> A ( -2 , 9 )

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3 years ago

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3 years ago

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Marrrta [24]

Answer:

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3 years ago

Write a polynomial functionſ of least degree that has rational coefficients, a leading coefficient of 1, and the zeros 1,3, and

otez555 [7]

Answer:

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Step-by-step explanation:

First of all, if 2 - i is a zero, then so is 2 + i

All you need to do now is multiply (x - 1)(x - 3)(x - 2 + i)(x - 2 - i)

I will do part of that multiplication

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So, now (x - 1)(x - 3)( $x^{2}$ - 4x + 5) = $x^{4} - 8x^{3} + 24x^{2} - 32x + 15$

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3 years ago

Find all solutions in the interval [0, 2π). (sin x)(cos x) = 0

Deffense [45]

$\bf sin(x)cos(x)=0\implies 
\begin{cases}
sin(x)=0\\
\measuredangle x=sin^{-1}(0)\\
\measuredangle x=0\ ,\ \pi \\
----------\\
cos(x)=0\\
\measuredangle x=cos^{-1}(0)\\
\measuredangle x=\frac{\pi }{2}\ ,\ \frac{3\pi }{2}
\end{cases}$

3 0

3 years ago

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