# If f(x) = x^6 - 3x^4 admits extremes in x = a, x=b and x=c then: a) a+b+c=1 b) abc =1 c)...

## Question:

If {eq}f(x) = x^6 - 3x^4{/eq} admits extremes in {eq}x = a{/eq}, {eq}x=b{/eq} and {eq}x=c{/eq} then:

a) {eq}a+b+c=1 {/eq}

b) {eq}abc =1 {/eq}

c){eq}a^2+b^2+c^2 = 4 {/eq}

d) {eq}a^2-b^2-c^2 = 1{/eq}

## Extremes

The ends of a function of a variable are the points where the first derivative is annulled and also it changes sign around that point.

We can also check the second derivative and depending on the sign of it, we classify the extremes.

If the second derivative evaluated at the point is positive, then the point is a local minimum, instead if the second derivative evaluated at the point is negative, then the point is a maximum.

Let f be a function twice derivable in an interval {eq}(a,b) {/eq} that contains the point {eq}x_0 {/eq} such that {eq}f '(x_0)= 0 \,\,\, \textrm {and} \,\,\, f ''(x_0) \neq 0 {/eq} And the second derivative is continuous in {eq}x_0 {/eq} It is true that:

1. If {eq}f ''(x_0) < 0 {/eq} then {eq}x_0 {/eq} is a local maximum point of the function f.

2. If {eq}f ''(x_0) > 0 {/eq} then {eq}x_0 {/eq} is a local minimum point of the function f.

## Answer and Explanation:

Step 1. Calculate the first derivative of the function.

{eq}f(x) = x^6 - 3x^4\\ f'(x) = 6x^5 -12x^3 {/eq}

Step 2. Find where the first derivative is canceled

{eq}f'(x) = 6x^5 -12x^3=0\\ 6x^3(x^2-2)=0\\ x=0 \,\,\,\,\,\, x=\pm \sqrt 2 {/eq}

The function admits extremes in {eq}x=0 \,\,\,\,\,\, x=\pm \sqrt 2 {/eq}

Is fulfilled c) {eq}a^2+b^2+c^2 = 4 {/eq} 