Check the solution to "Given h(x)=−2x2+x+1, find the absolute maximum value over..."

urca3m403

Answered question

2021-12-04

Given

h (x) = - 2 x^{2} + x + 1

, find the absolute maximum value over the interval [-3,3].

Gladys Fasching

Beginner2021-12-05Added 11 answers

Step 1
To determine the absolute maximum minimum.
Step 2
Given:

h (x) = - 2 x^{2} + x + 1

[-3,3]
Step 3
Compute the critical points on the equation as,

h (x) = - 2 x^{2} + x + 1

\frac{d}{d x} h (x) = \frac{d}{d x} (- 2 x^{2} + x + 1)

h^{'} (x) = \frac{d}{d x} (- 2 x^{2}) + \frac{d}{d x} (x) + \frac{d}{d x} (1)

h'(x)=-4x+1
h'(x)=0

\Rightarrow - 4 x + 1 = 0

\Rightarrow x = \frac{1}{4}

Step 4
Determine the value of the function at the points x = -3, x =1/4 and x=3.

h (- 3) = - 2 {(- 3)}^{2} + (- 3) + 1 = - 20

h (\frac{1}{4}) = - 2 {(\frac{1}{4})}^{2} + (\frac{1}{4}) + 1 = 1.125

h (3) = - 2 {(3)}^{2} + 3 + 1 = - 14

The minimum value of the function is -20 at x = -3.
Step 5
Thus, the absolute minimum value of the function is -20 at x=-3.

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