QuestionFind the limit of $\frac{f(x+h)-f(x)}{h}$ for $f(x)=x^{2}+1$ as $h \neq 0$ .

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=x^{}+1$ . We need to find the value of $\frac{f(x+h)-f(x)}{h}$ where $h \neq0$

STEP 2

First, we need to find the value of $f(x+h)$ . We can do this by replacing $x$ with $(x+h)$ in the function $f(x)$ .
$f(x+h) = (x+h)^{2}+1$

STEP 3

Now, expand the square in the above expression.
$f(x+h) = x^{2}+2xh+h^{2}+1$

STEP 4

Now, we have the values of $f(x+h)$ and $f(x)$ . We can substitute these values into the expression $\frac{f(x+h)-f(x)}{h}$ .
$\frac{f(x+h)-f(x)}{h} = \frac{(x^{2}+2xh+h^{2}+1)-(x^{2}+1)}{h}$

STEP 5

implify the numerator in the above expression.
$\frac{f(x+h)-f(x)}{h} = \frac{2xh+h^{2}}{h}$

STEP 6

Now, divide each term in the numerator by $h$ .
$\frac{f(x+h)-f(x)}{h} =2x+h$ So, the value of $\frac{f(x+h)-f(x)}{h}$ is $2x+h$ .

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QuestionFind the limit of $\frac{f(x+h)-f(x)}{h}$ for $f(x)=x^{2}+1$ as $h \neq 0$ .

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=x^{}+1$ . We need to find the value of $\frac{f(x+h)-f(x)}{h}$ where $h \neq0$

STEP 2

First, we need to find the value of $f(x+h)$ . We can do this by replacing $x$ with $(x+h)$ in the function $f(x)$ .
$f(x+h) = (x+h)^{2}+1$

STEP 3

Now, expand the square in the above expression.
$f(x+h) = x^{2}+2xh+h^{2}+1$

STEP 4

Now, we have the values of $f(x+h)$ and $f(x)$ . We can substitute these values into the expression $\frac{f(x+h)-f(x)}{h}$ .
$\frac{f(x+h)-f(x)}{h} = \frac{(x^{2}+2xh+h^{2}+1)-(x^{2}+1)}{h}$

STEP 5

implify the numerator in the above expression.
$\frac{f(x+h)-f(x)}{h} = \frac{2xh+h^{2}}{h}$

STEP 6

Now, divide each term in the numerator by $h$ .
$\frac{f(x+h)-f(x)}{h} =2x+h$ So, the value of $\frac{f(x+h)-f(x)}{h}$ is $2x+h$ .

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QuestionFind the limit of f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h}hf(x+h)−f(x)​ for f(x)=x2+1f(x)=x^{2}+1f(x)=x2+1 as h≠0h \neq 0h=0.

Studdy Solution

STEP 1

Assumptions1. The function is f(x)=x+1f(x)=x^{}+1f(x)=x+1 . We need to find the value of f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h}hf(x+h)−f(x)​ where h≠0h \neq0h=0

STEP 2

First, we need to find the value of f(x+h)f(x+h)f(x+h). We can do this by replacing xxx with (x+h)(x+h)(x+h) in the function f(x)f(x)f(x).f(x+h)=(x+h)2+1f(x+h) = (x+h)^{2}+1f(x+h)=(x+h)2+1

STEP 3

Now, expand the square in the above expression.f(x+h)=x2+2xh+h2+1f(x+h) = x^{2}+2xh+h^{2}+1f(x+h)=x2+2xh+h2+1

STEP 4

STEP 5

implify the numerator in the above expression.f(x+h)−f(x)h=2xh+h2h\frac{f(x+h)-f(x)}{h} = \frac{2xh+h^{2}}{h}hf(x+h)−f(x)​=h2xh+h2​

STEP 6

Now, divide each term in the numerator by hhh.f(x+h)−f(x)h=2x+h\frac{f(x+h)-f(x)}{h} =2x+hhf(x+h)−f(x)​=2x+hSo, the value of f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h}hf(x+h)−f(x)​ is 2x+h2x+h2x+h.

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QuestionFind the limit of $\frac{f(x+h)-f(x)}{h}$ for $f(x)=x^{2}+1$ as $h \neq 0$ .

Assumptions1. The function is $f(x)=x^{}+1$ . We need to find the value of $\frac{f(x+h)-f(x)}{h}$ where $h \neq0$

First, we need to find the value of $f(x+h)$ . We can do this by replacing $x$ with $(x+h)$ in the function $f(x)$ .
$f(x+h) = (x+h)^{2}+1$

Now, expand the square in the above expression.
$f(x+h) = x^{2}+2xh+h^{2}+1$

implify the numerator in the above expression.
$\frac{f(x+h)-f(x)}{h} = \frac{2xh+h^{2}}{h}$

Now, divide each term in the numerator by $h$ .
$\frac{f(x+h)-f(x)}{h} =2x+h$ So, the value of $\frac{f(x+h)-f(x)}{h}$ is $2x+h$ .