Asked by Marc
From first principles (ie using the tangent slope method), find the slope of the following curves at the given value of x.
f(x)=2x^2−6x at x = 3
f(x)=2x^2-6x
f(x+h)= 2(x+h)^2-6(x+h)
=2x^2+4xh+2h^2-6x-6h
lim h-->0 f(x+h)-f(x)/h
=2x^2+4xh+2h^2-6x-6h - (2x^2-6x)
=4xh+2h^2-6h/h
=h(4x+2h-6)/h
=lim h -->0 = 4x+2h-6
4x+2(0)-6
=4x-6
at x=3
f(3)=4x-6
=4(3)-6
=6
Im I doing this correct?
f(x)=2x^2−6x at x = 3
f(x)=2x^2-6x
f(x+h)= 2(x+h)^2-6(x+h)
=2x^2+4xh+2h^2-6x-6h
lim h-->0 f(x+h)-f(x)/h
=2x^2+4xh+2h^2-6x-6h - (2x^2-6x)
=4xh+2h^2-6h/h
=h(4x+2h-6)/h
=lim h -->0 = 4x+2h-6
4x+2(0)-6
=4x-6
at x=3
f(3)=4x-6
=4(3)-6
=6
Im I doing this correct?
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