Question

Find the point(s) on the graph of y=(2x^4+1)(x-5) where the slope is one.
I think I started on the right path not sure. I took the derivative of y and set it to zero???

Answer 1

The value of the first derivative IS the slope of the tangent line so set y' = 1.

Answer 2

I meant set it equal to 1

Answer 3

Yeah that's what I did. I just have trouble solving it. I got 10x^4-40x^3+x-5=0. I was like how do I solve that. If you can help I would really appreciate it

Answer 4

You've got to take the derivative of that expression and set it = to one.

Answer 5

oops. the derivative is actually 10x^4 - 40 x^3 +1

Answer 6

I did take the derivative of the expression and set it equal to one. This is my work:
y=(2x^4+1)(x-5)
y'=(2x^4+1)(1)+(x-5)(8x^3+1)
simplify further and you get
y'= 10x^4-40x^3+x-4
then i set y'=1 and it gave me: 10x^4-40x^3+x-5=0

Answer 7

Maybe I'm not doing it right

Answer 8

You have the derivative of (2x^4 + 1) as 8x^3 + 1. The 1 is a constant.

The derivative should simply be 8x^3.

Answer 9

ahhh. That's why it is always good to check and recheck your work. Thanks. I'm going to rework it and see what happens

Answer 10

ok so now I got 10x^4-40x^3+1=1

Answer 11

Bring one to the other side to get 10x^4-40x^3=0

Answer 12

I know you can take out a 10x^3 to get:10x^3(x-4)=0. So how would I go about solving this please. Is the quadratic formula feasible? I didn't think it was

Answer 13

\[10x ^{3} (x-4)=0\]

That gives you rwo locations where the slope of the tangent line = 1.

Answer 14

so do U solve for 10x^2 and x-4?

Answer 15

How did you get two locations? I got only one. x=4

Answer 16

My points are (4,-513). Don't know if it is right, but that's what I am sticking with as of right now

Answer 17

The other point is x=0 (0,-5)