Calculus experts?

Question

Can u help me! I am kinda stuck on this problem.

Unless otherwise specified, the domain of the function f is assumed to be the set of all real numbers x for which f(x) is a real number.

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 absolute minimum value of f. (show the analysis that leads to your conclusion)

c) find the x-coordinate of each point of inflection on the graph of f.
(show the analysis that leads to your conclusion)

Answer 1

Okay, so 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).
The equation of a line is y=mx+b
The slope at x= 2
f’(x)= 12x^3 + 3x^2 - 42x
f’(2)=96+12-84=24
so m=24 we have to find b
Substituting point (2, -28) into is y=24x+b we have
-28=48+b so b= -76
Finally if my arithmetic is correct the tangent line is y=24x-76

b) find absolute minimum value of f. (show the analysis that leads to your conclusion)
In general as people before me correctly stated the relative maxima and minima of a function can be found by finding the roots of the second derivative f”(x)=0. However thesese are relative values the obsolute value need to be found by trial.
Earlier we found that f’(x)= 12x^3 + 3x^2 - 42x
By finding the roots or when the tangent of the function equal to zero we compute potential candidates for maxima and minima.
x1=0
x2=14/8=7/4
x3=-16/8=-2
So f”(x)=36x^2+ 6x – 42
Let’s test
f”(0)= - 42
f”(7/4)= 78.75
f”(-2)=-90
It seems that the only maximum is at x=0
so f(0)=0

the maximum is at (0,0)

c) find the x-coordinate of each point of inflection on the graph of f.
(show the analysis that leads to your conclusion)
Did I forget to mention that we have point of inflection at f”(x)=0?
So if f’(x)= 12x^3 + 3x^2 - 42x then
f'”(x)= 36x^2 + 6x – 42
or
f'”(x)= 6x^2 + x - 7=0
The roots are
x1=1
x2=-14/12=-7/6
and these are the points of inflection
(1,y1) and (-7/6,y2) oh y1 and y2 you can compute by substituting x1 and corresponding x2 into th eorifinal equation for f(x).
Have fun!
And do not forget that a point must have (x and y coordinates to be a point) !

Answer 2

a) Find f'(x). Plug in x = 2. the result will be the slope of the tangent line. Use this slope, along with the point (2, -28), to get the equation of the tangent line.

b). Solve f'(x) = 0. You should get 3 solutions. Plug those solutions into f''(x). If f''(x) >0, then the relative extrema is a minimum. Find all values for which f'(x) = 0 and f''(x) > 0. Of these values, the one that produces the least value for f(x) is your absolute minimum.

c) Solve f''(x) = 0. You should get two points.

Answer 3

1.y=3x^4+x^3-21x^2
y'=12x^3+3x^2-42x
at x=2 this would be
96+12-84=24
the equation of the tangent
y+28=24(x-2)
24x-y-74=0

2.setting y'=0
x(12x^2+3x-4)=0
x=0 or roots of 12x^2+3x-4=0
now you take over from here

Answer 4

a) df(x)/dx evaluated at p will give you the slope of the tangent line at f(p).

b) when df(x)/dx = 0, you have a local minimum or maximum, don't forget to check f(+/- infinity) and then take the minimum value.

c) when d(df(x)/dx^2 = 0 (or rather when it moves from possitive to negative or vice versa) you have a point of inflection