What is the minimum value of the function f(x) = 18x^4 + 16x^2 - 16

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We can use Calculus to determine the minimum value of the equation. We are required to find the values of` x` where the first derivative`f'(x)=0` because the slope of the line at maximum and minimum point is zero. Moreover, to determine if the point on the graph is a minimum point, the second derivative `f''(a)` must be positive to state that the concavity at that point is upward(the point `a` is the lowest point in the graph) assuming that `x=a` is the solution of `f'(x) = 0` Now, if we take the first derivative of the equation, we get

`f'(x) = d/dx[18x^4+16x^2-16]`

`f'(x) = 18*d/dx(x^4)+16*d/dx(x^2)-d/dx(16)` 

`f'(x)=18*4(x^(4-1))+16*2(x^(2-1))-0`

`f'(x) = 72x^3+32x`

Then,

`f'(x) = 72x^3+32x` `=0`

`x(72x^2+32)=0`

This shows that the only real solution is `x=0`

The second derivative will be

`f''(x)=d/dx(72x^3+32x)`

`f''(x) = 72*3(x^(3-1))+32(1)`

`f''(x) =216x^2+32`

Then,

 `f''(0) = 216(0)^2+32=32`

which is positive

Thus, it shows that the minimum point of the function

`f(x) = 18x^4+16x^2-16` has a minimum point at `x=0` .

At `x=0` ,

`f(0) = 18(0)^4+16(0)^2-16 = -16`

Therefore, the minimum point on the graph is at `(0,-16).`

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The minimum value of a function f(x) can be determined using calculus. The function has an extreme value at the point where the first derivative f'(x) = 0. In addition, if x = a is the solution of f'(x) = 0, f''(a) should be positive.

For the function f(x) = 18x^4 + 16x^2 - 16,

f'(x) = 72*x^3 + 32x

72*x^3 + 32x = 0

=> x*(72x^2 + 32) = 0

The only real solution of this equation is x = 0.

The second derivative f''(x) = 216x^2 + 32. At x = 0, f''(0) = 32 which is positive.

The function f(x) = 18x^4 + 16x^2 - 16 has a minimum at x = 0. The value of the function at x = 0 is -16.

The minimum value of f(x) = 18x^4 + 16x^2 - 16 is -16.

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