Calculus I, concavity?

Question

Find all intervalls on which the graph of the function is concave upward. f(x)= (x^2+1) / (x^2). For the first derivative I got (-2) / (x^4). For the second derivative I got 6x^2/x^5 simplified to 6/x^3. On the interval (-inf.,-1) f"(-2)=negative and (-1,0) f"(-1/2)= neg.so concavity is downward. On the interval (0,1) f"(1/2)= positive and (1,+ inf.) f"(2)= pos. so concavity is upward.

Can anyone verify my answer please? Thanks : )

Oops for the first derivative I did get -2/x^3

Answer 1

The first derivative is (-2)/(x^3), making the second derivative 6/(x^4).

(it's easier to find the first derivative if you rewrite f(x) = 1 + 1/(x^2).)

I think you are making a mistake in how you are using the quotient rule.

The second derivative is positive for all x not equal 0. (x=0 is not in the domain of the original function.) Therefore, the function is concave upward (0,infinity) and (-infinity,0).

Answer 2

You took your derivatives wrong. The first derivative is -2/(x^3), and the second is 6/(x^4). The graph is apparently concave up on (-inf, inf), because the second derivative is positive on that interval. If you have problems taking derivatives, try http://www.quickmath.com/ It should do most of them for you. Good luck!

Answer 3

I might be a little rusty, but I think you went wrong at your first derivative. I got f'(x) = (-2)/(x^3)

From there...
f''(x) = 6/(x^4) --> x =/= 0
If you take test points at x = +/- 1, you'll see that it's always concave up.

Answer 4

f(x) = 2x? - 5x³ + 5 f'(x) = 8x³ - 15x² f"(x) = 24x² - 30x (using 2d by-product to envision concavity and inflection factors) set the 2d by-product to 0 and remedy for x, the inflection factors 24x² - 30x = 0 6x(4x-5) = 0 6x = 0 x = 0 4x-5 = 0 4x = 5 x = 5/4 determining on any consider between -? and nil, 0 and 5/4, 5/4 and ? to attempt concavity. if it really is adverse it really is concave down and if it really is costive it really is concave up. therefore -? < -a million < 0 ? f"(-a million) > 0 so (-?,0) is concave up 0 < a million <5/4 ? f"(a million) < 0 so (0.5/4) is concave down 5/4 < 2 < ? ? f"(2) > 0 so (5/4,?) is concave up (-?,0) ? (5/4,?) concave up (0.5/4) is concave down inflection factors at x = 0 and x = 5/4

Answer 5

i know im only 18 but i college credit for calc 1-4 so...

dy/dx= -2/x^3
dy'/d'x= 6/x^4

dy'/d'x= 0= 6/x^4 gives u a critical number at x=0 because of non existance

so f"(neg)= pos therefore concave up
and f"(pos)= pos and also concave up

Answer 6

f(x)=(x^2+1)/x^2
f(x)=1+1/(x^2)
f'(x)=-2/(x^3)
f''(x)=6/(x^4)
so the value of x is undefine
netheir it is not upward or downward but it may graph like fraction graph.