Derivatives Exercises

Answer

• f(x) = x² + 1, a = 1 We need a point and a slope. Since a = 1 and f(1) = 2, the point we want is (1, 2). We found earlier that f ' (1) = 2. We want a line

y = mx + b

whose slope is f ' (1) = 2, so

y = 2x + b.

We want the point (1, 2) to be on the line, so

2 = 2(1) + b.

Solving, we find b = 0 so the equation of the tangent line is

y = 2x.

We can check this with a picture:

Answer

• f(x) = x² + 1, a = 0We need a point and a slope. Since a = 0 and f(0) = 1, the point we want is (0,1). We found earlier that f'(0) = 0. We want a line

y = mx + b

whose slope is f'(0) = 0, so

y = (0)x + b = b.

We want the point (0,1) to be on the line, so

1 = b.

This means the equation of the tangent line is

y = 1.

We can check this with a picture:

Answer

• f(x) = x³, a = 1

We need a point and a slope. Since a = 1 and f(1) = 1, the point we want is (1, 1). We found earlier that f ' (1) = 3, so we can use this to find the line

y = mx + b

whose slope is f ' (1) = 3, so

y = 3x + b.

We want the point (1,1) to be on the line, so

1 = 3(1) + b.

Solving, we find b = -2 so the equation of the tangent line is

y = 3x – 2.

We can check this with a picture:

Looks like the slipper fits.

Answer

•  f(x) = 1 – x², a = -1

We need a point and a slope. Since a = -1 and f(-1) = 0, the point we want is (-1, 0). We found earlier that f ' (-1) = 2. We want a line

y = mx + b

whose slope is f ' (-1) = 2, so

y = 2x + b.

We want the point (-1, 0) to be on the line, so

0 = 2(-1) + b.

Solving, we find b = 2 so the equation of the tangent line is

y = 2x + 2.

We can check this with a picture:

Answer

Start with a point and a slope. Since a = 1 and , the point we want is . We found earlier that . We want a line

y = mx + b

whose slope is , so

We also want the point  to be on the line, so

Solving, we find  so the equation of the tangent line is

Before all is said and done, let's confirm this with a picture:

Looks like a tangent line to us.

Answer

• f(x) = 2x + 3x², a = 4

Since a = 4, f(4) = 2(4) + 3(4)² = 56. Now we need to find f ' (4).

Now put this all into the formula

y = f(a) + f ' (a)(x – a)

to find

y = 56 + 26(x – 4).

This can be simplified to

y = 26x – 48.

Answer

• f(x) = 2x³, a = -2

Since a = -2, f(a) = f(-2) = 2(-2)³ = -16.

To find f ' (-2), we have

We'll leave it for you to check that

f(-2 + h) = 2h³ – 12h² + 24h – 16.

Then

We put this all into the formula

y = f(a) + f ' (a)(x – a)

to find

y = -16 + 24(x – (-2)) = -16 + 24(x + 2).

This can be simplified to

y = 24x + 32.

Answer

Since a = 1, f(a) = f(1) = 1.

We need to find f ' (1):

We put this all into the formula

y = f(a) + f ' (a)(x – a)

to findy = 1 + (-2)(x – 1)which can be simplified toy = -2x + 3.

Answer

• The line is tangent to f at 1, so a = 1. Since the point (1, 0.5) is on the graph of f, f(a) = 0.5. To find f ' (a) we find the slope of the tangent line to f at a.