Question

Determine the inputs that yield the minimum values for each function. Compute the minimum value in each case. H(x) = 4x^4 − 4x^2 + 163

Answer 1

Okay. Find the first and second derivative.
\[H'(x)=16x^3 -8x\]
\[H''(x)=48x^2-8\]
At S.P's (Stationary Points)
\[H'(x)=0\]
\[16x^3-8x=0\]
\[2x^3-x=0\]
\[x(2x^2-1)=0\]
\[x=0\]
\[H(0)=163\]
\[H''(0)<0\]
Therefore maximum when x=0
\[x^2=\frac{1}{2}\]
\[x=\pm \frac{1}{\sqrt{2}}\]

Answer 2

Try find the y values for x=+/-[1/sqrt(2)]

Answer 3

H(x) values

Answer 4

and then substitute the values into the second derivative to see if the second derivative is less or greater than zero. If it's greater than zero, that's the minimum for whatever H(x) you get from your original equation.

Answer 5

wait so how did u come up with the derivatives?

Answer 6

You don't know how to differentiate?

Answer 7

@teresagenn

Answer 8

because the way that i was doing it was by factoring out and finding perfect square thats why im a little confused like this:
\[4x ^{4} - 4x ^{2}+163 \rightarrow 4x(x ^{3}-x+\frac{ 1 }{ 4 }) +164-1 \rightarrow 4x(x-\frac{ 1 }{ 2})^{2} +163\] and then get the x from the vertex to plug in to get the max value but i guess this is not correct right?

Answer 9

Nope. This is meant to be calculus finding the minimum and maximum