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f(x)=x^3-5x^2+3x-7. Using calculus methods find points where the line tangent to the curve is horizontal. Please show work!
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find the derivative. horizontal tangents occur when the derivative is equal to zero. f'(x)=3x^2-10x+3 Setting equal to zero: 3x^2-10x+3=0 (3x-1)(x-3)=0 This is zero when either 3x-1=0 or x-3=0 so, x=1/3 or x=3 Plug these x values into f(x) to get your points (1/3, f(1/3)) and (3, f(3)) where the tangent is horizontal.
wow that makes so much sense thanks so much
np :)
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