Given: f'(x)=4x^3+4x+6. How many points of inflection will the graph of f(x) have?
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Find the second derivative... find the derivative of this (since this equation is already the first derivative) and set it equal to 0.. that will be where your point(s) of inflection will be.
for my second derivative i got f"(x)=12x^2+4 but I dont think there will be any points am I correct?
If you set your second derivative equal to 0, you will get a point.
Not the answer you are looking for? Search for more explanations.
Should I use the quadratic formula?
Second derivative =
12x + 4 = 0
12x = -4
x= - 1/3
So you will have 1 point of inflection.
You can check this by integrating the first derivative to find the original function.
f(x) would be x^4 + 2x + 6x. Graph this and you will see there is one point of inflection.
No the second derivative would be 12\[x^2\]+4
b/c arent I suppose to subtract 1 from my exponent? each time I take my derivative
Oops, yes. Oh I see now. I would say no points of inflection
12x^2 + 4 = 0
x^2 = -1/3
but squares of numbers cannot be negatives