Let g be the function give by g(x) = x^4 - 4x^3 + 6x^2 4x + k where k is constant. A.On what...

Question:

Let g be the function give by {eq}g(x) = x^4 - 4x^3 + 6x^2 4x + k {/eq} where k is constant.

A.On what intervals is g increasing? Justify your answer.

B.On what interval is g concave upward? Justify your answer.

C.Find the value of k for which g has 5 as its relative minimum. Justify your answer.

Monotony and concavity:

We are looking for the monotony of this function, so we will need to derive the function with respect to the variable 'x'.

Then, we need the concavity, so we will need the second derivative.

Note that the values where the derivative is possitive, the funciton increases, while the values where the derivative is negative, the funciton decreases.

Answer and Explanation:

First of all, we assume that the function given is,

{eq}g(x) = x^4 - 4x^3 + 6x^2 - 4x + k {/eq}

The derivative is,

{eq}g'(x) = 4x^3-12x^2+12x-4 {/eq}

We are looking for the values of 'x' where the derivative is zero,

{eq}4x^3-12x^2+12x-4 = 0 \Rightarrow x^3-3x^2+3x-1=0 \Rightarrow (x-1)^3 = 0 {/eq}

So the only critical point is x=1. Let's see the sign of the derivative,

{eq}g'(0) = -4 < 0 \\ g'(2) = 4 > 0 {/eq}

Which means that,

{eq}g(x) \text{ decreases in } (-\infty, 1) \text{ because the derivative is negative in that interval} \\ gx) \text{ increases in } (1, \infty) \text{ because the derivative is positive in that interval} {/eq}

For the concavity we calculate the second derivative,

{eq}g''(x) = 12x^2-24x+12 {/eq}

The points when the concavity can change are when the second derivative is zero, which are,

{eq}x^2-2x+1 = 0 \Rightarrow x=1 {/eq}

So again we check the signs,

{eq}g''(0)=12 >0 \\ g''(2) = -12 < 0 {/eq}

So g is concave up when x<1 and concave down when x>1.

The last step is to find k when g has 5 as relative minimum. That is not possible due to the fact that the only possible relative extreme is when x=1.


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Concavity and Inflection Points on Graphs

from Math 104: Calculus

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