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IrinaVladis [17]
3 years ago
15

\[f(x)=\sqrt{x ^4-16x ^{2}}\]

Mathematics
1 answer:
marusya05 [52]3 years ago
4 0
f ( x ) =  \sqrt{ x^{4} -16 x^{2} }
a ) The domain:
x^4 - 16 x² ≥ 0
x² ( x² - 16 ) ≥ 0
x² - 16 ≥ 0
x² ≥ 16
x ∈ ( - ∞, - 4 ] ∪ [ 4 , + ∞ )
b ) f ` ( x ) = \frac{1}{2 \sqrt{x ^{4}-16 x^{2}  } } *  ( x^{4}-16 x^{2} )` =
= ( 2 x³ - 16 x ) / √(x^4 - 16 x²)
c ) The slope of the tangent line at x = 5:
f ` ( 5 ) = ( 2 * 125 - 16 * 5 ) / √ ( 625 - 400 ) = 170 / 15 = 34 / 3
The slope of the line normal to the graph at x = 5:
m = - 3 / 34 
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