Question

6. The function f given by f(x) = 3x +17, is (a) strictly increasing on R (c) decreasing on R (b) strictly decreasing on R (d) Both (b) and (c) are correct​

Answer 1

Answer:

The correct answer is option (A)  

Step-by-step explanation:

Given function is,

$f(x)=3x+17$

$\Rightarrow f'(x)=3(1)+0=3>0$

Now since $f'(x)>0$, so the function is strictly increasing

Answer 2

Answer:

The function f given by $f(x)=3x+17$ is strictly increasing on $R$.

Step-by-step explanation:

Given function is $f(x)=3x+17$.

We need to find the nature of the function.

If the value of differentiation is greater than zero, it is strictly increasing.

If the value of differentiation is lesser than zero, it is strictly decreasing.

For finding it, we will differentiate the function first.

${f}'(x)=\frac{d}{dx}(3x+17)$

Differentiation of $3x$ is $3$.

Differentiation of a constant is always zero.

Therefore, ${f}'(x)=3$.

The value of differentiation is greater than zero.

${f}'(x)=3>0$

So the function $f$ is strictly increasing on $R$.