find the slope f(x)=square root of x+1 over square root of x+2 where x=4
m= 0, b= 1/6 where m is the slope and b is the y-intercept
is it possible for you to write up the solution?
\(\Huge\bf Welcome~to~ \color{#00B4ff}{Open}\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Huge\bf\color{#7cc517}{Study~!!}~\) My name is Stephen and if you have any questions, come to me. \[f(x) = \frac{\sqrt{x + 1}}{\sqrt{x + 2}}, x = 4\] Now, simply take x and plug it in \[f(4) = \frac{\sqrt{4 + 1}}{\sqrt{4 + 2}} = f(4) = \frac{\sqrt{5}}{\sqrt{6}}\] Now, simply solve the fraction.
hey its actually square root of x then +1 and square root of x then +2
Okay, so \[f(4) = \frac{ \sqrt{4} + 1}{ \sqrt{4} + 2}\] Alright! So what I did was take our x + 1 and x + 2 and simply replace the x value with 4 (because we know x = 4). Now, what do you think the square root of 4 is? Remember, a square root is a number that produces a specified quantity when multiplied by itself. "7 is a square root of 49" "5 is a square root of 25" and so forth. So what is the square root of 2?
so the slope is 3 over 7? but don't we have to take the derivate of the function first?
\[\frac{ 5 }{ 6 }\]
i think you forgot to take the square root of 4
I'm glad you caught that. I didn't know you wanted the derivative. Are we simplying or is there something else? \[\frac{ 2 + 1 }{ 2 + 2 } = \frac{ 3 }{ 4}\]
oh it says find the slope of the tangent line to the curve f(x)=square root of x+1 over square root of x+2 is 25.7 at the point where x=4 it is true or false question and the answer is false i just wanted to know the right answer instead of 25.7
That would've been really helpful information Lol! I don't know calculus very well. We can consult @Directrix ; @Nnesha ; @eliassaab on this
oh anyways I appreciate your help
|dw:1418190970115:dw| this is your question
@jenn0117
rationalize it
|dw:1418191041639:dw|
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