What is the equation of the tangent line of #f(x)=(2x^3 - 1) / (x^2-2x) # at #x=3#?

Answer 1

To find the equation of the tangent line to the function ( f(x) = \frac{{2x^3 - 1}}{{x^2 - 2x}} ) at ( x = 3 ), follow these steps:

  1. Find the derivative of the function ( f(x) ) using the quotient rule.
  2. Evaluate the derivative at ( x = 3 ) to find the slope of the tangent line.
  3. Use the point-slope form of a line equation with the slope from step 2 and the point ( (3, f(3)) ) to determine the equation of the tangent line.

Derivative of ( f(x) ): [ f'(x) = \frac{{(2x^3 - 1)'(x^2 - 2x) - (2x^3 - 1)(x^2 - 2x)'}}{{(x^2 - 2x)^2}} ] [ f'(x) = \frac{{(6x^2)(x^2 - 2x) - (2x^3 - 1)(2x - 2)}}{{(x^2 - 2x)^2}} ] [ f'(x) = \frac{{6x^4 - 12x^3 - 4x^4 + 4x + 2x^3 - 2}}{{(x^2 - 2x)^2}} ] [ f'(x) = \frac{{2x^4 - 6x^3 + 4x - 2}}{{(x^2 - 2x)^2}} ]

Evaluate ( f'(3) ): [ f'(3) = \frac{{2(3)^4 - 6(3)^3 + 4(3) - 2}}{{(3^2 - 2(3))^2}} ] [ f'(3) = \frac{{162 - 162 + 12 - 2}}{{(9 - 6)^2}} ] [ f'(3) = \frac{{10}}{{9}} ]

Using the point-slope form: [ y - f(3) = f'(3)(x - 3) ] [ y - \frac{{2(3)^3 - 1}}{{(3)^2 - 2(3)}} = \frac{{10}}{{9}}(x - 3) ] [ y - \frac{{52}}{{7}} = \frac{{10}}{{9}}(x - 3) ]

Thus, the equation of the tangent line at ( x = 3 ) is: [ y = \frac{{10}}{{9}}x - \frac{{4}}{{63}} ]

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Answer 2

#y=-158/9*x+211/3#

Let

#u=2x^3-1#
#v=x^2-2x#

Then

#f(x)=u/v#

Using the quotient rule, which can be found in all good google searches:

#f'(x)=(u'v-uv')/v^2#
#u'=6x#
#v'=2x-2#
#f'(x)=([6x(x^2-2x)]-[(2x^3-1)(2x-2)])/(x^2-2x)^2#

Soz, I cannot be bothered simplifying this...

#f'(x)# is the gradient function, substituting in any x value will give us the gradient of the tangent at that point.

So let's sub in x=3:

#f'(3)=([6*3(3^2-2*3)]-[(2*3^3-1)(2*3-2)])/(3^2-2*3)^2#
#rArrf'(3)=([18(9-6)]-[(54-1)(6-2)])/(9-6)^2=(18*3-53*4)/9=-158/9#

Now we want the equation of the tangent, which is a straight line of the form:

#y=mx+c#
We know #m=-158/9# but we need to solve for #c#. We need a point on the line to solve it, so sub in #x=3# into the original equation to find the y value:
#f(3)=(2*3^3-1)/(3^2-2*3)=(54-1)/(9-6)=53/3#

So the coordinates of our point on the line are (3,53/3). Sub this point into the straight line equation to solve for c:

#y=-158/9*x+c#
#rArr53/3=-158/9*3+c#
#rArr53/3=-158/3+c#
#rArrc=53/3+158/3=211/3#

So the equation of the tangent is:

#y=-158/9*x+211/3#
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Answer 3

To find the equation of the tangent line of f(x) at x=3, we need to find the derivative of f(x) and evaluate it at x=3. The derivative of f(x) can be found using the quotient rule. After finding the derivative, we can substitute x=3 into the derivative to find the slope of the tangent line. Finally, we can use the point-slope form of a line to write the equation of the tangent line.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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