Why we also check at interval boundaries also...finding critical points with F' = 0

Finding Critical points with F' = 0 , Why it doesn't gives boundaries as critical point IF they are either MAX/MIN...

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17 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.

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.

0

edited
Jun 21, 2020
by sanjaysharmarose

Why we are not calculating f ''(x) which is:

f ''(x) = 4 , and f ''(x) >0 at 0,1/2 and 2.

https://gateoverflow.in/2664/gate1995-25a (This is the method I am talking about.)

0

16 votes

in such type of questions it's better to know the nature of curve

**General case** :- Let $f(x)=ax^{2}+bx+c$ where $a\neq0$

** a)** concave or convex

1) if $a>0$ then curve will open up (convex nature)

2) if $a<0$ then curve will open down (concave nature)

**b) Nature of roots**

Let $D=b^{2}-4ac$

1) if $D>0$ , then our curve will intersect $x-axis$ at two different points (means two different real roots)

2) if $D=0$ , then our curve will touch $x-axis$ at one point (means two same real roots)

3) if $D<0$ , then our curve won't touch or cut $x-axis$. Means either it will be completely above the $x-axis$(when $a>0$) or it will be entirely below the $x-axis$ (when $a<0$).Hence imaginary roots.

**Original question**

$f(x)=2x^{2}-2x+6$

1) $a=2$ , so $a>0$ means curve will be open up nature

2) $D=-44$ , so $D<0$ means our curve will be entirely above $x-axis$ , mean imaginary roots.

$\frac{\mathrm{d} y}{\mathrm{d} x}=4x-2=0$ , so at $x=\frac{1}{2}$ , we have Max/Min.

although from graph we can directly it would be minima at $x=\frac{1}{2}$ , yet another way is to know the nature of second derivative which is $\frac{\mathrm{d} (4x+2)}{\mathrm{d} x}=4$ , so $+ve$ means minima at $x=\frac{1}{2}$

So at that level from graph we can directly say , either graph will have maxima in $[0,2]$ at $x=0$ or $x=2$ , so just calculate the value at those points and see which one is max.

5 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]

\(f'(x) = 4x - 2\)

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]