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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 | 245 views

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 (34k points)
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 (163 points)
It can also be written as $2((x-\frac{1}{2})^2 + \frac{11}{4})$ and then check for max val of $(x - 1/2)$

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 (3.3k points)

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