# Thread: find equation of tangent line to graph at the function given value of x

1. ## find equation of tangent line to graph at the function given value of x

f(x)= -2x^(1/2)+x^(3/2), x=4.

So I know we need a point, and a slope.

f(4)= -2(sqrt4) + (sqrt16) with an index of 3.

f(4)= -4+ 2(sqrt2) with an index of 3.

so are point is (4, -4+ 2(sqrt2) with an index of 3)

now we need to find f'(4) to find the slope which is : limit of x approaching 4 = f(x)- f(4)/ (x-4)

I feel like I am not doing something right. I do not think I can find a conjugate with this problem.

2. ## Re: find equation of tangent line to graph at the function given value of x

$f(x) = -2x^{1/2} + x^{3/2}$

$\dfrac{df}{dx} = -x^{-1/2} + \dfrac 3 2 x^{1/2}$

$f(4) = 4$

so we have the point $(4,4)$ on the line.

$\left . \dfrac{df}{dx}\right|_{x=4}=\dfrac 5 2$

so we have a line with slope $\dfrac 5 2$ passing through the point $(4,4)$

$y-4=\dfrac 5 2 (x-4)$

$y = \dfrac 5 2 x -6$

3. ## Re: find equation of tangent line to graph at the function given value of x

Originally Posted by math951
f(x)= -2x^(1/2)+x^(3/2), x=4.

So I know we need a point, and a slope.

f(4)= -2(sqrt4) + (sqrt16) with an index of 3.
What in the world does this mean? Surely you know (you use it in the next line) that sqrt(4)= 2 and I assume that by "index of 3" you mean "to the third power". No, $\displaystyle 2^3$\displaystyle is not "sqrt{16}" (which is 4 by the way). $\displaystyle 4^{3/2}$ can be done either as $\displaystyle (4^3)^{1/2}= (64)^{1/2}= 8$ or $\displaystyle (4^{1/2})^3= 2^3= 8$.

Yes, for any differentiable function, $\displaystyle f'(a)= \lim_{x\to a}\frac{f(x)- f(a)}{x- 1}$, by definition, but by the time you are doing
problems like this you should have learned at least that $(x^b)'= bx^{b- 1}$ as well as that (f(x)+ g(x))'= f'(x)+ g'(x). That is, $\displaystyle (-2x^{1/2}+ x^{3/2})'= -2(1/2)x^{-1/2}+ (3/2)x^{1/2}= -x^{-1/2}+ (3/2)x^{1/2}$ which, at x= 4, is $\displaystyle -(1/2)+ (3/2)(2)= -1/2+ 3= 5/2$

f(4)= -4+ 2(sqrt2) with an index of 3.

so are point is (4, -4+ 2(sqrt2) with an index of 3)

now we need to find f'(4) to find the slope which is : limit of x approaching 4 = f(x)- f(4)/ (x-4)

I feel like I am not doing something right. I do not think I can find a conjugate with this problem.[/QUOTE]