# 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

Relevance
• Anonymous

(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.

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

. . -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