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Find the minimum or maximum value of the function f(x)=8x2+x−5. Give your answer as a fraction.

Find the minimum or maximum value of the function f(x)=8x2+x−5. Give your answer as-example-1

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Answer

Minimum value of the function = (-41/8)

Step-by-step explanation

The minimum or maximum of a function occurs at the turning point of the graph of the function.

At this turning point, the first derivative of the function is 0.

The second derivative of the function is positive when the function is at minimum and it is negative when the function is at maximum.

f(x) = 8x² + 2x - 5

(df/dx) = 16x + 2

At minimum or maximum point,

16x + 2 = 0

16x = -2

Divide both sides by 16

(16x/16) = (-2/16)

x = (-1/8)

Second derivative

f(x) = 8x² + 2x - 5

(df/dx) = 16x + 2

(df²/d²x) = 16 > 0, that is, positive.

So, this point is a minimum point.

f(x) = 8x² + 2x - 5

f(-1/8) = 8(-1/8)² + 2(-1/8) - 5

= 8 (1/64) - (1/4) - 5

= (1/8) - (1/4) - 5

= (1/8) - (2/8) - (40/8)

= (1 - 2 - 40)/8

= (-41/8)

Hope this Helps!!!

User Ashil John
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