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What is the maximum value of the function $f(x) = 2x^2 - 2x + 6$ in the interval $\left[0,2 \right]$?

  1. 6
  2. 10
  3. 12
  4. 5.5

 

asked in Calculus by Veteran (66.1k points) 1148 2197 2522 | 416 views

3 Answers

+7 votes
Best answer
Answer: B

For $f(x)$ to be maximum

$f'(x) = 4x - 2 = 0 \\ \implies x = \frac{1}{2}$

So at $x = \frac{1}{2}, f(x)$ is an extremum (either maximum or minimum).

$f(2) = 2(2)^2 - 2(2) + 6 = 8 - 4 + 6 = 10$

$f\left(\frac{1}{2}\right) = 2{\frac{1}{2}}^2 - 2\frac{1}{2} + 6 = 5.5$, so $x = \frac{1}{2}$ is a mimimum.

$f(0) = 6$.

So, the maximum value is at $x = 2$ which is $10$ as there are no other extremum for the given function.
answered by Veteran (35.4k points) 52 204 346
+2 votes
Here \(f(x) = 2x^2 - 2x + 6\)
\(f'(x) = 4x - 2\)
critical point is \(4x - 2 =0\)  at \(x=1/2\)
if we see the number line of f(x) then first it is decreasing at interval ($-\infty$ to 1/2 ) and increasing at interval (1/2 to $\infty$)
so at \(x=1/2 \) it is getting its minimum value.
So the maxima for interval [0, 2] can be at either at \(x=0\) or \(x=2\)  because it was decreasing till 1/2 and started increasing after 1/2.
\(f(0) = 6\)
\(f(2) = 10\)
So the maximum value is 10 for the interval [0, 2]
answered by (261 points) 1 3 6
It can also be written as $2((x-\frac{1}{2})^2 + \frac{11}{4})$ and then check for max val of $(x - 1/2)$
0 votes

answer is (b)

f'(x)=4x-2=0⇒x=1/2

so for getting maximum value from the function  in inteval[0,2] we bhav to put

f(0)=6

f(1/2)=5.5

f(2)=10

so 10 is maximum 

answered by Loyal (4.4k points) 6 39 85


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