# f(x)= 3x^(4)+ x^(3)-21x^2?

Let f be the function given by f(x)= 3x^(4)+ x^(3)-21x^2

a. write an equation of the line tangent to the graph of f at the point (2,-28).

b. Find all relative minimums for F(x)

c. Find the x coordinate of each point of inflection on the graph of f. Justify

### 5 Answers

- Anonymous1 decade agoBest Answer
(a) The tangent line is y = [f '(x)]x + b. So first, you need to find the derivative of the function.

f '(x) = 3(4x^3) + 3x² - 21(2x)

f '(x) = 12x^3 +3x² - 42x

Next, plug in the x value to get the slope.

f '(2) = 12(2)^3 +3(2)² - 42(2)

f '(2) = 96 +12 - 84

f '(2) = 24

Then, find b.

y = [f '(2)]x + b

-28 = (24)(2) + b

b + 48 = -28

b = -76

So your tangent line is y = 24x - 76

(b) First, find all the critical numbers.

f '(x) = 12x^3 +3x² - 42x

0 = 12x^3 +3x² - 42x

0 = 3x(4x² + x - 14)

0 = 3x(4x - 7)(x + 2)

x = 0, 7/4, -2

The critical numbers are 0, 7/4, and -2. Now you have to find the intervals where f '(x) is positive and the intervals where f '(x) is negative.

Draw your number line:

<---|---|----|------>

. . -2 0 7/4

The dots don't mean anything. Please don't include them on the paper. They are there because without them, the line would look like this on the computer:

<---|---|----|------>

-2 0 7/4

Anyway, back to the problem. You need to find whether f '(x) is positive or negative for the intervals (-∞, -2), (-2, 0), (0, 7/4), and (7/4, ∞). Plug in a test x value for each interval.

Starting with (-∞, -2):

f '(-3) = 12(-3)^3 +3(-3)² - 42(-3)

f '(-3) = -324 - 27 + 126

f '(-3) = -225

Mark the interval you just determined as negative.

. -

<---|---|----|------>

. . -2 0 7/4

Keep doing this until you know how it goes for all intervals.

f '(-1) = 12(-1)^3 +3(-1)² - 42(-1)

f '(-1) = -12 + 9 + 42

f '(-1) = 39

. - . +

<---|---|----|------>

. . -2 0 7/4

f '(1) = 12(1)^3 +3(1)² - 42(1)

f '(1) = 12 + 3 - 42

f '(1) = -27

. - . + . -

<---|---|----|------>

. . -2 0 7/4

f '(2) = 12(2)^3 +3(2)² - 42(2)

f '(2) = 96 +12 - 84 (look familiar?)

f '(2) = 24

. - . + . - . +

<---|---|----|------>

. . -2 0 7/4

If f '(x) is negative before the critical number and positive after, f(x) has a relative minimum for that critcal number. If f '(x) is positive before the critical number and negative after, f(x) has a relative maximum for that critcal number. If the sign is the same before and after the critical point, there is no relative extrema at that point.

Basically, the x values at which f(x) has a relative minimum in this case are -2 and 7/4. So just plug into the equation of f(x) to find the points.

f(-2) = 3(-2)^4 + (-2)^3 - 21(-2)²

f(-2) = 48 - 8 - 84

f(-2) = -44

f(7/4) = 3(7/4)^4 + (7/4)^3 - 21(7/4)²

f(7/4) = 7203/256 + 343/64 - 1029/16

f(7/4) = -7889/256

The relative minimums are (-2, -44) and (7/4, -7889/256).

(c) First find the second derivative.

f '(x) = 12x^3 +3x² - 42x

f "(x) = 12(3x²) + 3(2x) - 42

f "(x) = 36x² + 6x - 42

Now find x when f "(x) = 0

0 = 36x² + 6x - 42

0 = 6(6x² + x - 7)

0 = 6(6x + 7)(x - 1)

x = -7/6, 1

And that's your answer.

I am certain because we know from part (b) that these points lie between the relative minima and the relative maximum. The process for finding those two x values we found reveal x values where a point of inflection is possible, but there is always a point of inflection when you shift from a relative minimum to a relative maximum and vice-versa.

- nicotLv 43 years ago
i in my opinion choose childrens could supply up posting retardedly simplistic homework that may no longer be able to be defined without in easy terms offering you with the recommendations. Ask your mum and dad that would fairly assist you you. in the event that they are actually no longer able to determine those out then.... my god...

- druckLv 43 years ago
i in my view choose infants can supply up posting retardedly simplistic homework that isn't be waiting to be defined with out basically providing you with the ideas. Ask your moms and dads that could extremely assist you you. in the event that they're now unable to establish those out then.... my god...

- How do you think about the answers? You can sign in to vote the answer.