Question

Let g(x)=6x-x^2+14sqrt(x)-15. Find the slope of the tangent line to the graph of g(x) at x=4.

Answer 1

\(\bf g(x)=6x-x^2+14\sqrt{x}-15\qquad \textit{use the power rule}\\ \quad \\
\textit{keep in mind that}\quad \sqrt{x}\implies \large x^{\frac{1}{2}}\)

Answer 2

I'll see if I can find that in my notes... thanks.

Answer 3

But what do I do with the g(x)=4?

Answer 4

once you find the derivative, you replace x = 4 :)
g(x) = f'(x)

Answer 5

If I do the power rule I get... 6-2x+7x^(-1/2). Is that right? And what do I do from there if it is?

Answer 6

welp... that came out off...

well..

once you have the derivative , just set x = 4, f'(4)

Answer 7

I got it wrong? That's not the derivative?

Answer 8

you're correct, that's what I got

Answer 9

What did I do wrong? 6x -> 6. -x^2 -> -2x. 14x^(1/2) -> 7x^-1/2, no?

Answer 10

\(\bf g(x)=6x-x^2+14\sqrt{x}-15\\ \quad \ g'(x) = 6-2x+7x^{\frac{1}{2}}\qquad then\quad
g'(4)\)

Answer 11

hmm forgot the -

Answer 12

Oh. So now just plug in 4? And thats the answer?

Answer 13

\(\bf g(x)=6x-x^2+14\sqrt{x}-15\\ \quad \\
g'(x) = 6-2x+7x^{-\frac{1}{2}}\qquad then\quad
g'(4)
\)

Answer 14

So 3/2?

Answer 15

yeap, the derivative of a function is the "slope function" for the function

if you want to know the value of the slope at "x", just set "x" to that

Answer 16

yes, I got 3/2

Answer 17

Thank you so much. I really needed to get this right. It's for a takehome quiz and I'm doing terrible in the class. Hopefully this helps. :P Thanks a lot, man.

Answer 18

yw