Sean asked in Science & MathematicsMathematics · 1 month ago

How do I solve this derivative question?

I have no idea how to make a graph like that.

Relevance
• 1 month ago

f(x) = (3x+1) if x</= 1 or (-2x+6) if x >/= 1

h(x) = f(g(x)

h(x) = (3(2x+1) + 1 if (2x+1)</= 1

==> h(x) = 6x + 4 if 2x </= 0

or h(x) = (-2(2x+1) +6) if (2x+1)>/= 1

==> h(x) = (-4x +4) if 2x >/= 0

if x = 1 then 2x> 0

==> h(x) = -4x +4

==> h'(x) = - 4

==> h'(-1) = -4

• 1 month ago

The rise slope and fall slope are different.

E non existent.

• ?
Lv 7
1 month ago

Nonexistent .............ANS (Option E)

• Dixon
Lv 7
1 month ago

Since g(x) = 2x + 1

f[g(x)] = f(2x + 1)

Then

h(x) = f(2x + 1)

We are given f(x) on the graph, so we can transform it to f(2x + 1) as shown in the sketches below.

Then by inspection, at x = 1 the gradient is the slope of the downward line segment.

h'(1) = (y2 - y1)/(x2 - x1)

h'(1) = -6/1.5

h'(1) = -4 . . . . . . . . ANSWER

We can also reason that if h'(1) does map onto the downward line segment, then adding 1 has no effect on the slope, but multiplying by 2 will make the slope twice as steep. And the original slope is -6/3 = -2, so that is OK.

[Added: Some answers say there isn't a single value for the slope at x = 1 but the graph in the question is for f(x) not h(x) ]

• fooks
Lv 7
1 month ago

1.

Refer to the graph of the function f.

The slopes of f(x) at x = 1 are different in the two line segments.

When 0 < x ≤ 1: f(x) = m₁x + a

When 1 ≤ x ≤ 4: f(x) = m₂x + b, where m₁ ≠ m₂

When 0 < x ≤ 1:

h(x) = f[g(x)]

h(x) = f(2x + 1)

h(x) = m₁(2x + 1) + a

h(x) = 2m₁x + (m₁ + a)

h'(x) = 2m₁

h'(1) = 2m₁

When 1 ≤ x ≤ 4:

Similarly, h'(1) = 2m₂

As the values of h'(1) are different (2m₁ ≠ 2m₂) in the two line segments, h'(1) is nonexistent.