Question

if f(x) = x sqrt 5 - x, write the equation of the tangent and normal line to the graph at the point where x = 1

Answer 1

Is it \(\large x\sqrt{5-x}\) or \(\large x\sqrt{5} - x\)?

Answer 2

the first one

Answer 3

First thing to note is that when \(\large x=1\), you have \(x\sqrt{5-x} = 1\cdot \sqrt{4} = 2\). Thus, we're to find the equation of the tangent and normal lines passing through the point \(\large (1,2)\).

So what you need to do now is compute the derivative of \(\large x\sqrt{5-x} = x(5-x)^{1/2}\). We see that
\[\large \begin{aligned}\frac{d}{dx}(x(5-x)^{1/2}) &= x\cdot((5-x)^{1/2})^{\prime} + (x)^{\prime}\cdot (5-x)^{1/2} \\ &= x\cdot \frac{1}{2}(5-x)^{-1/2}\cdot (-1) + (5-x)^{1/2}\end{aligned}\]
What do you get when you plug in \(\large x=1\)? This value will be the slope of the tangent line. Then you get the slope of the normal line by taking the opposite reciprocal. At this point, you can determine the tangent and normal line equations.

Does this make sense? :-)

Answer 4

let me see what i get

Answer 5

i got tangent= y-2=-1/2x+1
Normal= y-2=2x+1

Answer 6

When you plug in \(\large x=1\) into the derivative, you should get \(-\dfrac{1}{4} + 2 = \dfrac{7}{4}\). So the slope of the tangent is \(\dfrac{7}{4}\) and the slope of the normal is \(-\dfrac{4}{7}\).

Does it make sense why this is the case?

Answer 7

yeah makes total sense thank you so much