root function analysis

[natanson.meital]

Here is the function [tex]y=x*\sqrt{a+b*x^2}[/tex]
the tangent equation of the function at [tex]x=\sqrt{5}[/tex]
is [tex]x+2y=5*\sqrt{5}[/tex]

A. Find a and b.
B. Find the function definition range.
C. Find the minimum & maximum points.

We tried to derive the function but we didn't succeed.
Please help us to finish the question.
Thanks.

[Math Tutor]

the tangent equation of a function:
[tex]y - y_0 = f'(y_0)(x - x_0)[/tex]

[tex]x_0 = \sqrt{5}[/tex]

[tex]y_0 = \sqrt{5}\sqrt{a + 5b}[/tex]


and we also know that [tex]x + 2y = 5\sqrt{5}[/tex] here we have [tex]x = \sqrt{5}[/tex] and [tex]2y = 2\sqrt{5}\sqrt{a + 5b}[/tex]

Continue, please I am waiting...
We have a system of 2 equations

[natanson.meital]

hello!

I understood that when [tex]x=\sqrt{5}[/tex]
then we can put it in the function and the tangent equation to get the Y value in 2 ways and to compare between them.

When i put [tex]x=\sqrt{5}[/tex] in the function then i get [tex]y=\sqrt{5}*\sqrt{a+5b}[/tex]
And when i put [tex]x=\sqrt{5}[/tex] in the tangent then i get [tex]y=2*\sqrt{5}[/tex]
And when i compare between them, i get [tex]4=a+5b[/tex].

But i didn't understood where i get from the second equation that you talked about.

Can u explain me please?

Thanks a lot.

[Math Tutor]

Find the tangent equation by the formula
[tex]y - y_0 = f'(y_0)(x - x_0)[/tex]

This will be the second equation
when
[tex]x = \sqrt{5}[/tex]
[tex]x = 2\sqrt{5}[/tex]

[natanson.meital]

Thanks for your answer.

I thought to get the second equation in this way:
To derivate the tangent equation and the function equation and to compare between them in [tex] x=\sqrt{5}[/tex]
Because the two derivations give me the incline of the function and the tangent in [tex] x=\sqrt{5}[/tex] and it has to be the same.

Is this way right?

[Math Tutor]

yes
Could you write the result, please.

[sajid121]

To derivate the tangent equation and the function equation and to compare between them in [tex]x=\sqrt{5}[/tex]
Because the two derivations give me the incline of the function and the tangent in [tex]x=\sqrt{5}[/tex] and it has to be the same.

[Rock73]

Here is the function y=x∗a+b∗x2−−−−−−−−√y=x∗a+b∗x2
the tangent equation of the function at x=5–√x=5
is x+2y=5∗5–√x+2y=5∗5

A. Find a and b.
B. Find the function definition range.
C. Find the minimum & maximum points.

i could not understand about how to solve?