# Check if the function f(x)= 10x^2 - 2x + 5 as extreme values.

Asked on by zarvamea

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The extreme values  of f(x)  is  the solution  for x in f'(x) = 0.

So we  find f'(x). Abd equate it to zero and solve for x.

f(x) = 10x^2-2x +5 .

f'(x) = (10x^2-2x+5)'

f'(x) = 20x -2.

f'(x) = 0 gives: 20x-2 = 0

20x= 2

x = 2/20 = 1/10.

So f(x) has an estreme value at x = 1/10.

f" (x) = (20x-2)' = 20 is positive.  So f"(1/10) > 0.

Therefore f(x) has a minimum for x = 1/10.

Therefore f(x)  = 10x^2-2x+5  has  the minimum f(1/10) = 10(1/10^2) -2/10+5 = 1/10 -1/5 +5 =  4.9  at x = 1/10.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The given function has a single extreme value, namely a minimum value (bcause the coefficient of x^2 is positive). The minimum value is represented by the vertex of the parable, whose expression is 10x^2 - 2x + 5.

The coordinates of the vertex are: V(xV , yV):

xV = -b/2a

yV = -delta/4a

delta = b^2 - 4ac

We'll identify the coefficients a,b,c:

a = 10

b = -2

c = 5

Now, we'll determine the coordinates of the vertex:

xV = -(-2)/2*10

xV = 1/10

xV = 0.1

yV = -(4 - 200)/4*10

yV = 196/4*10

yV = 49/10

yV = 4.9

The function has just one extreme point and it's coordinates are: (0.1 , 4.9).

Another manner to verify the existence and the number of extreme points of a function is to differentiate the function.

We'll differentiate f(x).

f'(x) = (10x^2 - 2x + 5)'

f'(x) = 20x - 2

We'll calculate the roots of f'(x):

f'(x) = 0

20x - 2 = 0

We'll divide by 2:

10x - 1 = 0

x = 1/10

The function has an extreme point for any root of the derivative.

Since the derivative has just one root, the function will have just a single extreme point.

f(1/10) = 10/100 - 2/10 + 5

f(1/10) = 1/10 - 2/10 + 5

f(1/10) = -1/10 + 5

f(1/10) = (-1+50)/10

f(1/10) = 49/10

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