Question

Let f(x)=2x^4-16x^2+4

Use the first derivative test to find the value(s) of x for which has a relative maximum and list them in increasing order.
9there should be 3 answers)

Answer 1

(there should be 3 answers)

Answer 2

X=?

Answer 3

They can be max or min, three extrema are present.

Answer 4

yes you but i want to know the maxima

Answer 5

Well firstly you have to use the derivative \[8x ^{3}-32x\] you can see an x is common to both so you get \[x(8x ^{2}-32)\] and then you factorise the bracket and set it equal to zero. Shouldn't be too difficult. :)

Answer 6

then what do i do after i factorise?

Answer 7

you now have your x values, so you can sort them, lowest to highest, you should get x=-2,0,2 from the top of my head

Answer 8

those are your maximas and minimas

Answer 9

\[\frac{d}{dx} (2x ^{4} - 16x ^{2} + 4) = 8x ^{3} - 32x = 0 => 8 x (x ^{2} - 4) = 0\]

Hence, we can say, the extrema are present @ x = 0, or x = 2 or x = -2

Now,
\[\frac{d^{2}}{dx^{2}} (2x ^{4} - 16x ^{2} + 4) = 24x ^{2} - 32\]

Hence @ x = 0, \[\frac{d^{2}}{dx^{2}} (2x ^{4} - 16x ^{2} + 4) = - 32\]
Hence @ x = 2, \[\frac{d^{2}}{dx^{2}} (2x ^{4} - 16x ^{2} + 4) = 64\]
Hence @ x = -2, \[\frac{d^{2}}{dx^{2}} (2x ^{4} - 16x ^{2} + 4) = 64\]

Hence the maxima is present at x = 0 and minima are present @ x= 2 and x = -2