Compute Difference Quotient

[nycmath]

Precalculus
David Cohen
Edition 3
Chapter 3, S2ction 3.1

Compute the difference quotient for the function.

See attachment.

[Math Tutor]

The algebra is perfect. With [tex]f(x)=-x^{2}+6x-2[/tex]:

[tex]\frac{f(x+h)-f(x)}{h}=\frac{-2xh-h^{2}+6h}{h}=-2x-h+6[/tex]

One small correction to the labels at the bottom, though. Two of the three are right, the last one is not:

[tex]-2x-h+6[/tex] is the slope of the secant line through [tex](x,\,f(x))[/tex] and [tex](x+h,\,f(x+h))[/tex], and that is the same thing as the average rate of change of [tex]f[/tex] on [tex][x,\,x+h][/tex]. So slope = average rate of change, yes.

But it is not the derivative yet - notice it still contains [tex]h[/tex]. The derivative is what is left after you let the second point slide into the first:

[tex]f'(x)=\lim_{h\to 0}\left(-2x-h+6\right)=-2x+6[/tex]

So the difference quotient is the average rate of change, and the derivative is the instantaneous rate of change.

Sanity check at [tex]x=3[/tex]: the difference quotient gives [tex]-h[/tex], and [tex]f'(3)=0[/tex]. That matches the graph - [tex]x=3[/tex] is the vertex of the parabola, where the tangent is horizontal.

[Eigenvalue]

Yes, this is correct. The value becomes the derivative when the limit of h approaches 0

[nycmath]

The algebra is perfect. With [tex]f(x)=-x^{2}+6x-2[/tex]:

[tex]\frac{f(x+h)-f(x)}{h}=\frac{-2xh-h^{2}+6h}{h}=-2x-h+6[/tex]

One small correction to the labels at the bottom, though. Two of the three are right, the last one is not:

[tex]-2x-h+6[/tex] is the slope of the secant line through [tex](x,\,f(x))[/tex] and [tex](x+h,\,f(x+h))[/tex], and that is the same thing as the average rate of change of [tex]f[/tex] on [tex][x,\,x+h][/tex]. So slope = average rate of change, yes.

But it is not the derivative yet - notice it still contains [tex]h[/tex]. The derivative is what is left after you let the second point slide into the first:

[tex]f'(x)=\lim_{h\to 0}\left(-2x-h+6\right)=-2x+6[/tex]

So the difference quotient is the average rate of change, and the derivative is the instantaneous rate of change.

Sanity check at [tex]x=3[/tex]: the difference quotient gives [tex]-h[/tex], and [tex]f'(3)=0[/tex]. That matches the graph - [tex]x=3[/tex] is the vertex of the parabola, where the tangent is horizontal.

Your reply is double perfect. Thank you.

[nycmath]

Yes, this is correct. The value becomes the derivative when the limit of h approaches 0

Thank you so much. The limit of h tends to 0 but never quite reaches 0. Yes?