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Svet_ta [14]
3 years ago
10

Graph the function f(x) = x3 – 5x2 + 6x using the graphing calculator. What are the solutions to the related equation? Check all

that apply.
–3
–2
–1
0
1
2
3

Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
3 0

the solutions to the related equation are 0,2,3 .

<u>Step-by-step explanation:</u>

Here we have , function  f(x) = x3 – 5x2 + 6x . Graph of this function is given below  . We need to find What are the solutions to the related equation . Let's find out:

Solution of graph means the  value of x at which the value of f(x) or function is zero . We can determine this by seeing the graph as at what value of x does the graph intersect or cut x-axis !

At x = 0 .

From the graph , at x=0 we have f(x) = 0 i.e.

⇒ f(x) = x^3- 5x^2 + 6x

⇒ f(0) = 0^3- 5(0)^2 + 6(0)

⇒ f(0) =0

At x = 2 .

From the graph , at x=2 we have f(x) = 0 i.e.

⇒ f(x) = x^3- 5x^2 + 6x

⇒ f(0) = 2^3- 5(2)^2 + 6(2)

⇒ f(0) =0

At x=3 .

From the graph , at x=3 we have f(x) = 0 i.e.

⇒ f(x) = x^3- 5x^2 + 6x

⇒ f(0) = 3^3- 5(3)^2 + 6(3)

⇒ f(0) =0

Therefore , the solutions to the related equation are 0,2,3 .

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Answer:

  a) d(sinh(f(x)))/dx = cosh(f(x))·df(x)/dx

  b) d(cosh(f(x))/dx = sinh(f(x))·df(x)/dx

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Step-by-step explanation:

To do these, you need to be familiar with the derivatives of hyperbolic functions and with the chain rule.

The chain rule tells you that ...

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The attached table tells you the derivatives of the hyperbolic trig functions, so you can answer the first three easily.

__

a) sinh(u)' = sinh'(u)·u' = cosh(u)·u'

For u = f(x), this becomes ...

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__

b) After the same pattern as in (a), ...

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__

c) Similarly, ...

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__

d) For this one, we need the derivative of sech(x) = 1/cosh(x). The power rule applies, so we have ...

  sech(x)' = (cosh(x)^-1)' = -1/cosh(x)²·cosh'(x) = -sinh(x)/cosh(x)²

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Now, we will use this as above.

  sech(4x+2)' = -sech(4x+2)·tanh(4x+2)·(4x+2)'

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_____

Here we have used the "prime" notation rather than d( )/dx to indicate the derivative with respect to x. You need to use the notation expected by your grader.

__

<em>Additional comment on notation</em>

Some places we have used fun(x)' and others we have used fun'(x). These are essentially interchangeable when the argument is x. When the argument is some function of x, we mean fun(u)' to be the derivative of the function after it has been evaluated with u as an argument. We mean fun'(u) to be the derivative of the function, which is then evaluated with u as an argument. This distinction makes it possible to write the chain rule as ...

  f(u)' = f'(u)u'

without getting involved in infinite recursion.

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