Question

an equation for the line tangent to the graph of the differentiable function f at x=3 is y=4x+6. which of the following statements must be true?

Answer 1

Answer:

C. II and III

Step-by-step explanation:

See attachment for complete question

Given

$f(x) = 4x + 6$  differentiable at $x = 3$

Start by representing y as $f(x)$;

$f(x) = 4x + 6$

Solving (I): $f(0) = 3$

Substitute 0 for x in $f(x) = 4x + 6$

$f(0) = 4 * 0 + 6$

$f(0) = 0 + 6$

$f(0) = 6$

I is not true

Solving (II): $f(3) = 18$

Substitute 3 for x in $f(x) = 4x + 6$

$f(3) = 4 * 3 + 6$

$f(3) = 12 + 6$

$f(3) = 18$

II is true

Solving (III): $f'(3) = 4$

First , we have to differentiate  $f(x)$

$f(x) = 4x + 6$

$f'(x) = 1 * 4x^{1 - 1} + 0$

$f'(x) = 1 * 4x^{0} + 0$

$f'(x) = 1 * 4 + 0$

$f'(x) = 4$

Then, substitute 3 for x in $f'(x) = 4$

$f'(3) = 4$

III is true