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Slav-nsk [51]
3 years ago
7

What is the local maximum value of the function? g (x)=x^4-5x^2+4

Mathematics
1 answer:
Neporo4naja [7]3 years ago
5 0

g(x)=x^4-5x^2+4 \\ g'(x)=4x^3-5 \cdot 2x^1+0=4x^3-10x \\ \\ 4 x^{3} -10x=0 \\ 2x(2x^2-5)=0 \\ 2x^2-5=0~~~~and~~~2x=0 \\ 2x^2=5~~~~~~~~~~~~~~~~~x=0 \\ x^2=2.5 \\ x= \sqrt{2.5}  \\ x=1,58 ~~~~~~and ~~~~x= -1.58
__-____-1,58____+_____0_______-__1,58_____+____>x

Locol max: 0

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Tpy6a [65]

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Let's solve your equation step-by-step.

0.25x2−0.25x−10.5=0

For this equation: a=0.25, b=-0.25, c=-10.5

0.25x2+−0.25x+−10.5=0

Step 1: Use quadratic formula with a=0.25, b=-0.25, c=-10.5.

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