If f(x) = ax + b and f(0) = 2 and f '(0) = 1, then the value of f(2) is

Open in App

Solution

We have,
f(x) = ax + b
Differentiating both sides w.r.t. x, we get
f '(x) = a
Therefore, f '(0) = 1 means a = 1.
f(0) = b means b = 2.
Therefore, f(x) = x + 2.
f(2) = 2 + 2 = 4.

Suggest Corrections

Similar questions

Q. Let

f (\frac{x + y}{2}) = \frac{1}{2} [f (x) + f (y)]

for real x and y. If

f^{'} (0)

exists and equals

- 1

and

f (0) = 1

then the value of

f (2)

Let $f (x) = a x^{2} + b x + c$ where a,b, and c are constants. If f(x) takes it maximum value at $x = \frac{1}{3},$ then which of the following is necessarily true?

Q. If

f (x) = (1 + x)^{n}

, then the value of

f (0) + f^{'} (0) + \frac{f^{''} (0)}{2!} + \dots + \frac{f^{n} (0)}{n!}

Q. If 2 and 0 are the zeros of the polynomial

f (x) = 2 x^{3} - 5 x^{2} + a x + b

then find the values of a and b.
Hint f(2) = 0 and f(0) = 0.

Q. If

f (\frac{x + y}{2}) = \frac{f (x) + f (y)}{2} \forall x, y \in R

and

f^{'} : (0) = - 1, f (0) = 1,

then

f (2) = ?

Join BYJU'S Learning Program

If f(x) = ax + b and f(0) = 2 and f '(0) = 1, then the value of f(2) is

We have, f(x) = ax + b Differentiating both sides w.r.t. x, we get f '(x) = a Therefore, f '(0) = 1 means a = 1. f(0) = b means b = 2. Therefore, f(x) = x + 2. f(2) = 2 + 2 = 4.

We have,
f(x) = ax + b
Differentiating both sides w.r.t. x, we get
f '(x) = a
Therefore, f '(0) = 1 means a = 1.
f(0) = b means b = 2.
Therefore, f(x) = x + 2.
f(2) = 2 + 2 = 4.