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12.4 Derivatives  (Page 7/18)

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Graph of f(x) = x^(1/3) with a viewing window of [-3, 3] by [-3, 3].
The graph of   f ( x ) = x 1 3   has a vertical tangent at   x = 0.

Differentiability

A function   f ( x )   is differentiable at   x = a   if the derivative exists at   x = a , which means that   f ′ ( a )   exists.

There are four cases for which a function   f ( x )   is not differentiable at a point   x = a .

  1. When there is a discontinuity at   x = a .
  2. When there is a corner point at   x = a .
  3. When there is a cusp at   x = a .
  4. Any other time when there is a vertical tangent at   x = a .

Determining where a function is continuous and differentiable from a graph

Using [link] , determine where the function is

  1. continuous
  2. discontinuous
  3. differentiable
  4. not differentiable

At the points where the graph is discontinuous or not differentiable, state why.

Graph of a piecewise function that has a removable discontinuity at (-2, -1) and is discontinuous when x =1.

The graph of f ( x ) is continuous on ( −∞, −2 ) ∪ ( −2 , 1 ) ∪ ( 1, ∞ ). The graph of f ( x ) has a removable discontinuity at x = −2 and a jump discontinuity at x = 1. See [link] .

Graph of the previous function that shows the intervals of continuity.
Three intervals where the function is continuous

The graph of is differentiable on ( −∞, −2 ) ∪ ( −2, −1 ) ∪ ( −1, 1 ) ∪ ( 1, 2 ) ∪ ( 2, ∞ ) . The graph of f ( x ) is not differentiable at x = −2 because it is a point of discontinuity, at x = −1 because of a sharp corner, at x = 1 because it is a point of discontinuity, and at x = 2 because of a sharp corner. See [link] .

Graph of the previous function that not only shows the intervals of continuity but also labels the parts of the graph that has sharp corners and discontinuities. The sharp corners are at (-1, -1) and (2, 3), and the discontinuities are at (-2, -1) and (1, 1).
Five intervals where the function is differentiable
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Determine where the function   y = f ( x )   shown in [link] is continuous and differentiable from the graph.

Graph of a piecewise function with three pieces.

The graph of   f   is continuous on   ( − ∞ , 1 ) ∪ ( 1 , 3 ) ∪ ( 3 , ∞ ) .   The graph of   f   is discontinuous at   x = 1   and   x = 3.   The graph of   f   is differentiable on   ( − ∞ , 1 ) ∪ ( 1 , 3 ) ∪ ( 3 , ∞ ) .   The graph of   f   is not differentiable at   x = 1   and   x = 3.

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Finding an equation of a line tangent to the graph of a function

The equation of a tangent line to a curve of the function   f ( x )   at   x = a   is derived from the point-slope form of a line,   y = m ( x − x 1 ) + y 1 .   The slope of the line is the slope of the curve at   x = a   and is therefore equal to   f ′ ( a ) , the derivative of   f ( x )   at   x = a .   The coordinate pair of the point on the line at   x = a   is   ( a , f ( a ) ) .

If we substitute into the point-slope form, we have

The point-slope formula that demonstrates that m = f(a), x1 = a, and y_1 = f(a).

The equation of the tangent line is

y = f ' ( a ) ( x − a ) + f ( a )

The equation of a line tangent to a curve of the function f

The equation of a line tangent to the curve of a function   f   at a point   x = a   is

y = f ' ( a ) ( x − a ) + f ( a )

Given a function   f ,   find the equation of a line tangent to the function at   x = a .

  1. Find the derivative of   f ( x )   at   x = a   using   f ′ ( a ) = lim h → 0 f ( a + h ) − f ( a ) h .
  2. Evaluate the function at   x = a .   This is   f ( a ) .
  3. Substitute   ( a , f ( a ) )   and   f ′ ( a )   into   y = f ' ( a ) ( x − a ) + f ( a ) .
  4. Write the equation of the tangent line in the form   y = m x + b .

Finding the equation of a line tangent to a function at a point

Find the equation of a line tangent to the curve   f ( x ) = x 2 − 4 x   at   x = 3.

Using:

f ' ( a ) = lim h → 0 f ( a + h ) − f ( a ) h

Substitute   f ( a + h ) = ( a + h ) 2 − 4 ( a + h )   and   f ( a ) = a 2 − 4 a .

  f ′ ( a ) = lim h → 0 ( a + h ) ( a + h ) − 4 ( a + h ) − ( a 2 − 4 a ) h           = lim h → 0 a 2 + 2 a h + h 2 − 4 a − 4 h − a 2 + 4 a h Remove parentheses .           = lim h → 0 a 2 + 2 a h + h 2 − 4 a − 4 h − a 2 + 4 a h Combine like terms .           = lim h → 0 2 a h + h 2 − 4 h h           = lim h → 0 h ( 2 a + h − 4 ) h Factor out  h .           = 2 a + 0 − 4 f ′ ( a ) = 2 a − 4 Evaluate the limit .   f ′ ( 3 ) = 2 ( 3 ) − 4 = 2

Equation of tangent line at   x = 3:

y = f ' ( a ) ( x − a ) + f ( a ) y = f ' ( 3 ) ( x − 3 ) + f ( 3 ) y = 2 ( x − 3 ) + ( − 3 ) y = 2 x − 9
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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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