Some help me with these two questions please!!!!!!!
1 answer:
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9514 1404 393
Answer:
(√5)/2
Step-by-step explanation:
Of the several ways I can think of to do this, using a graphing calculator is about the easiest. It shows the minimum to be ...
f(1) = √1.25 = (√5)/2
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Using the distance formula, you have ...
f(x) = √((x -(-2))² +((x² +2)-5/2)²)
f(x) = √(x² +4x +4 +x⁴ -x² +1/4) = √(x⁴ +4x +17/4)
The minimum is found where the derivative is zero.
f'(x) = (2x³ +2)/√(x⁴ +4x +17/4) = 0
x³ = -1 . . . . . f'(x) is zero when the numerator is zero
x = -1 . . . . . cube root
Then the minimum value of f(x) is ...
f(-1) = √(x⁴ +4x +17/4) = √((-1)⁴ +4(-1) +17/4) = √(1 -4 +17/4) = √(5/4)
f(-1) = (√5)/2 . . . . minimum value of f(x)
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The graph shows f²(x) in red and its minimum of 1.25 = 5/4. The curve (x, x²+2) and the point (-2, 5/2) are also shown, for reference. (The slope of the curve at x=-1 is -2, and the normal to the curve at that point has slope 1/2. The normal goes through the point (-2, 5/2), consistent with f(x) being a minimum at x=-1.)
