Question

The function f is defined by f(x)= sqrt (25-x^2) for -5 less than/equal to x less than/equal to 5
a. write an equation for the line tangent to the graph of f at x=-3

Answer 1

take the derivative, replace \(x\) bu \(-3\)

Answer 2

\[f(x)=\sqrt{25-x^2}\]
\[f'(x)=\frac{-x}{\sqrt{25-x^2}}\]

Answer 3

i am going to guess you get \(\frac{3}{4}\) but you can check it

Answer 4

what i mean is that the slope will be \(\frac{3}{4}\)

Answer 5

it is clear that this function is the upper half of the circle with radius 5 centered at the origin?

Answer 6

It's not specified in prompt if that's what you mean, but i guess it would be