For what value of $h$ is the minima equal to $-5$

Question

I have some questions regarding this exercise.

$f_h(x)=x^4-hx^2$

"For what value of $h$ is the minima equal to $-5$"

What I would have thought is correct:

$f_h'(x)=(4x^3-2hx)=0$

$=x(4x^2-2h)=0$

$x=0$ and $x=\pm\sqrt{\frac{h}{2}}$

Then, I would say okay, well if $h=10$ then this would satisfy the question. Meaning the answer is $h=10$. But for some reason, the answer sheet states the following steps instead:

Take $x=\pm\sqrt{\frac{h}{2}}$ and input into the function:

$f(-\sqrt{\frac{h}{2}})=f(\sqrt{\frac{h}{2}})$

= ${\frac{h^2}{2}}=5$ Therefore, $h=\sqrt{10}$.

I don't quite get how that way worked. They figured out the slope, put it into the original equation and equaled it to $-5$, how does that give them the required value.

Would appreciate some explanation, thanks in advance.

Answer 1

$x=0$ and $x=\pm\sqrt{\frac h2}$

Then, I would say okay, well if $h=10$ then this would satisfy the question.

What? Why? I don't understand this at all.

....

I don't quite get how that way worked. They figured out the slope,

Not quite, this isn't a line so there is no single slope. They figured out what the derivative was and this gives a formula for the slope (of the tangent line) at any given $x$.

What the actually figured out was the three points where the slope will be equal to $0$. At $x = 0$ then the slope of the tangent line is $0$. ANd $x= \sqrt{\frac h2}$ the slope of the tangent line is $0$ and at $x= -\sqrt{\frac h2}$ the slope of the tangent line is $0$. At all other values of $x$ the slope of the tangent line is not $0$.

put it into the original equation

Not quite. The slope at $x_0$ is $0$. They did not put the slope into the original equation. They put the value of $x$ where $f'(x) = 0$ back into the original equation.

So when $f'(x) = 0$ then $f(x)$ is at an extreme value.

and equaled it to −5.

Because we were told $f(x) = -5$ when $x$ is such that $f(x)$ is a minimum value.

how does that give them the required value.

It's a basic observation, that extrema points (max, mins, and saddles) of continuous functions can only and always appear when the slope of the tangent line is $0$.

So .... if $(x_0, f(x_0))$ is a local minimum, then $f'(x_0) = 0$. And if $f'(x_0) =0$ then $x_0 = 0$ or $x_0 =\pm \sqrt{\frac h2}$.

So either $(0, f(0)) = (0, 0)$ is the minimum value $(0, -5)$ (which is impossible as $0 \ne -5$.

or $(\sqrt{\frac h2}, f(\sqrt{\frac h2})) = (\sqrt{\frac h2}, \frac {h^2}4 - h* \frac h2)= (\sqrt{\frac h2}, -\frac {h^2}4)$ is the minimum value of $(\sqrt{\frac h2}, -5)$

Or $(-\sqrt{\frac h2}, f(-\sqrt{\frac h2}) = (-\sqrt{\frac h2}, \frac {h^2}4 - h* \frac h2)= (-\sqrt{\frac h2}, -\frac {h^2}4)$ is the minimum value of $(-\sqrt{\frac h2}, -5)$

As $f(0) = 0 > -5$ that is neither the minimum nor is it $-5$.

So $f(\pm \sqrt{\frac h2} = -\frac {h^2}4 = -5$ are the two local minima.

So what is the $h$ that will make this true?

So we solve for $-\frac {h^2}4 = -5$.

Which is $h=\pm \sqrt{20}$. But for $x = \pm \sqrt {\frac h2}$ to be defined we must have $h\ge 0$ so $h = \sqrt{20}$.

Answer 2

The minimum of the function should be $-5$ . It means if $f'(a) = 0$ then $f(a) = -5$ . Here we have $a = \pm \sqrt {h/2}$

Answer 3

you have combined things that should be written separately. For a chosen $h,$ the minimum value occurs when $x = \pm \sqrt {h/2}.$ Let's see, the original function was $f_h(x)=x^4 - h x^2.$ Alright $$ f_h( \sqrt {h/2} ) = \frac{h^2}{4} - \frac{h^2}{2} = - \frac{h^2}{4} \; . $$ They want this to be $-5,$ so we need $$ - \frac{h^2}{4} = -5, $$ $$ h^2 = 20, $$ but $h$ must be positive so $$ h = \sqrt{20} = 2 \sqrt 5$$

Always wise to check. We have $\frac{h}{2} = \sqrt 5,$ so that the $x$ value we are using is $5^{1/4}.$ Then the function value becomes $$ 5 - 2 \sqrt 5 \sqrt 5 = 5 - 10 = -5 $$

Note that $\sqrt{20} \approx 4.47 \; , \;$ while $5^{1/4} \approx 1.49534878$ is pretty close to $\frac{3}{2}$