Question

If x be real, find the maximum value of 3-6x-8x2 and the corresponding value of x.

Answer 1

HEY!!

Let f(x) = -8x2 - 6x + 3

f(-3) = -72 + 18 + 3 = -51

f(-2) = -32 + 12 + 3 = -17

f(-1) = -8 + 6 + 3 = 1

f(0) = 3

f(1) = -8 - 6 + 3 = -11

f(2) = -32 - 12 + 3 = -41

f(3) = -72 - 18 + 3 = -87

So, as x → -, f(x) → - and as x → , f(x) → -

Hence, maximum value of f(x) is 3 at x = 0

Answer 2

The maximum value is 33/8 and corresponding value of x = - 3/8

Given :

3 - 6x - 8x²

To find :

The maximum value and corresponding value of x

Solution :

Step 1 of 2 :

Write down the given function

Let f(x) be the given function

Then f(x) = 3 - 6x - 8x²

Step 2 of 2 :

Find maximum value and corresponding value of x

$\sf{Here \: \: f(x) = 3 - 6x - 8{x}^{2} }$

Differentiating both sides with respect to x two times we get

$\sf{f'(x)= - 6 - 16x}$

and

$\sf{f''(x)= - 16\: }$

Now for extremum value of f(x) we have

$\sf{f '(x)= 0\: }$

$\implies \sf{ - 6 - 16x = 0\: }$

$\implies \sf{ - 16x = 6\: }$

$\implies \sf{ x = \frac{6} {-16} }$

$\implies \sf{ x = -\frac{3} {8} }$

Now

$\displaystyle \sf f'' \bigg( - \frac{3}{8} \bigg) = - 16 < 0$

∴ f(x) has maximum at x = - 3/8

The maximum value

$\displaystyle \sf = f\bigg( - \frac{3}{8} \bigg)$

$\displaystyle \sf = 3 - 6 \times \bigg( - \frac{3}{8} \bigg) - 8 \times {\bigg( - \frac{3}{8} \bigg)}^{2}$

$\displaystyle \sf = 3 + \frac{18}{8} - \frac{9}{8}$

$\displaystyle \sf = \frac{24 + 18 - 9}{8}$

$\displaystyle \sf = \frac{33}{8}$

Hence maximum value is 33/8 and corresponding value of x = - 3/8

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