# Thread: Can anyone help me figure this out. Critical points.

1. ## Can anyone help me figure this out. Critical points.

(a) Let a > 0 be a constant. Find all critical points of f (x) = x + a√x.

This is my work so far

f'(x)=1+(a/2)(x^(-1/2))
0=1+(a/2)(x^(-1/2))
-1 = (a/2)(1/sqrt(x))
-1=a/(2sqrt(x))
(2)(-1)=(a/(2sqrt(x)))(2)
-2=a/sqrtx
(a)-1/2=sqrtx/a (a)
-a/2=sqrtx
(-a/2)^2=(sqrtx)^2
a/4=x

I calculated the Critical point as x=a/4

(b) Use derivatives to show that f is increasing and its graph is concave down for all x > 0.

Not sure how to do this part.

2. Originally Posted by dptrimble
(a) Let a > 0 be a constant. Find all critical points of f (x) = x + a√x.

This is my work so far

f'(x)=1+(a/2)(x^(-1/2))
0=1+(a/2)(x^(-1/2))
-1 = (a/2)(1/sqrt(x))
-1=a/(2sqrt(x))
(2)(-1)=(a/(2sqrt(x)))(2)
-2=a/sqrtx
(a)-1/2=sqrtx/a (a)
-a/2=sqrtx
(-a/2)^2=(sqrtx)^2
a/4=x

I calculated the Critical point as x=a/4

(b) Use derivatives to show that f is increasing and its graph is concave down for all x > 0.

Not sure how to do this part.

The function's derivative is $\displaystyle 1+\frac{a}{2\sqrt{x}}=1+\frac{a}{2}x^{-\frac{1}{2}}$

Thus, $\displaystyle f'(x)>0\, \forall{x}$ and thus it is increasing.
Now evaluate the second derivative and prove it is always negative.

Tonio