What is the function? f(x)=2x^2-3x+1

To satisfy the hypotheses of the Mean Value Theorem a function must be continuous in the closed interval and differentiable in the open interval.

Step-by-step explanation:

As f(x)=2x3−3x+1 is a polynomial, it is continuous and has continuous derivatives of all orders for all real x, so it certainly satisfies the hypotheses of the theorem.

To find the value of c, calculate the derivative of f(x) and state the equality of the Mean Value Theorem:

dfdx=4x−3

f(b)−f(a)b−a=f'(c)

f(x)x=0=1

f(x)x=2=3

Hence:

3−12=4c−3

and c=1.

0= 2x^2 - 3x + 1

2x^2 -3x+1=0

x= -(-3)+radical(-3)^2 -4×2×1 over 2×2 =

x=3+ radical 9-8 over 4=

x=3 + radical 1 over 4

x= 3+1 over 4

x=3-1 over 4

x=1

x =1 over 2

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .