Relative maxima or minima?

Question

Okay. Normally i know how to find the maxima or minima but for some reason this one is throwing me off.. Can i get some help?

y=5x^2 + x

Please and Thank you!

Answer 1

well, first take the derivative.

y' = 10x + 1

find the zeros

0 = 10x + 1
-1 = 10x
-1/10=x

find a value less than -1/10 for y' and a value greater than -1/10 for y'.

if the first value is negative and the second is positive, then it is a minimum.

if the first value is positive and the second value is negativem then it is a maximim.

once you find what it is, plug in -1/10 for x and solve y=5x^2+x

that will give you your y coodinate.

and your answer will be (-1/10, {whatever y is})

Answer 2

y = 5x^2 + x + 0
Using Calculus.
dy/dx = 10x + 1
at max/min dy/dx = 0
therefore 10x + 1 = 0
10x = -1
x = -1/10 the 'x' coordinate of the min'n point.
as the coefficient of x^2 is '+'ve then the curve is 'cup' shaped and x = -1/10 is a minimum.
and y = 5(-1/10)^2 + -1/10
y = 5/100 - 1/10 = 1/20 - 1/10
y = 1/20 - 2/20 = -1/10

So coords (x,y) = (-1/10, -1/10) at minimum.

Answer 3

y' = 10x + 1
0 = 10x + 1
10x = -1
x = -1/10

This is minimum.

y = 5(-1/10)^2 - 1/10
y = 1/20 - 1/10
y = -1/20

Local minium: (-1/10,-1/20)