Question

Find the tangent line to the curve y =
√x at x = 4
(1) y=x/4+1
(3) y = 5x + 7
(2) y = 3x -2
(4) y = 4x+8​

Answer 1

Answer:

option 1

Explanation:

see the attached image

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Answer 2

Equation of tangent line to the curve at

x=4 is $\bf y = \frac{x}{4} + 1$.

Option 1 is correct.

Explanation:

Given:

• A curve.
• $y = \sqrt{x}$

To find:

• Find the equation of tangent at x=4,
• (1) y=x/4+1
• (2) y = 5x + 7
• (3) y = 3x -2
• (4) y = 4x+8

Solution:

Concept to be used:

• Equation of tangent on curve y:$\bf y - y1 = m(x - x1)$, here $\bf m = \frac{dy}{dx}$ is first derivative at given point and refer as slope of line.

Step 1:

Find slope of equation.

Find the first derivative of y.

$\frac{dy}{dx} = \frac{1}{2 \sqrt{x} }$(apply power rule of differentiation)

put x=4

$\frac{dy}{dx} = \frac{1}{2 \sqrt{4} }$

or

$\bf m = \pm\frac{1}{4}$

We have to discard the negative value of slope, as curve is define for positive values.

Thus,

Slope of tangent line is 1/4.

Step 2:

Find the point (x1,y1).

Put x= 4 in the curve.

$y = \sqrt{4}$

so,

$\bf y = \pm2$

Points are $\bf (4,2)$ and $\bf (4, - 2)$

The curve is not define for negative values, so we have to discard (4,-2) point.

Thus,

Point of intersection of tangent line and curve is (4,2).

Step 3:

Find the equation of tangent.

Tangent line passes through (4,2) having slope 1/4.

$y - 2 = \frac{1}{4} (x - 4)$

or

$y = \frac{x}{4} - \frac{4}{4} + 2$

or

$y = \frac{x}{4} - 1 + 2$

or

$\bf y = \frac{x}{4} + 1$

Thus,

One equation of tangent line to the

curve at x=4 is $\bf y = \frac{x}{4} + 1$.

Option 1 is correct.

Learn more:

1) find the equation of tangent & normal to the curve 4x^2+9y^2=40 at point(1,2)

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